<|instruction|>
Solve the below problem. The solution should start with an abridged problem statement. Then key observations. Then full solution based on the observations. Then C++ and Python implementations with comments.

517. Cornerless Tiling
Time limit per test: 0.25 second(s)
Memory limit: 262144 kilobytes
input: standard
output: standard



An cornerless tiling of an mx n rectangle is such tiling of this rectangle with 1x 2 and 2x 1 dominoes that no four dominoes share a corner.

For example, here are the two possible cornerless tilings of 4x 4 square:


How many cornerless tilings of mx n rectangle are there?

Input
First and only line of the input file contains two integers m and n, 1 ≤ n, m ≤ 1000.

Output
In the only line of the output file write the sought number of tilings.

Example(s)
sample input
sample output
4 4
2

<|response|>
1. Abridged problem statement
Given an m×n grid (1≤m,n≤1000), count the number of ways to tile it with 1×2 and 2×1 dominoes so that no point of the grid is a corner shared by four dominoes. If the total number of cells is odd, the answer is 0. Print the exact count (no modulo).

2. Key observations
- A domino covers two cells ⇒ if m·n is odd, answer = 0.
- We can swap m,n so that n≤m, reducing cases.
- For n=1, there's exactly one way (all dominoes lie along the length).
- For n=2, a classic 1D DP: let dp[i] = ways to tile 2×i strip under the "no four-corner" rule. You can extend by one column with a vertical domino (dp[i−1]) or by two horizontal domino pairs that step back three columns (dp[i−3]).
- For n>2, there are only two "cornerless" tiling patterns of an n×n block (rotations of each other). By concatenating such blocks (or their n×(n±2) variants), you reduce the problem to 1D DP on columns.
  - If n is odd, no column can be fully vertical throughout ⇒ a single-state DP f[i] with transitions f[i]=f[i−(n−1)] + f[i−(n+1)], and final answer = 2·f[m] (two global orientations).
  - If n is even, some columns may be fully vertical. We need two DP states per width i:
     • dp[i][0] = ways with column i fully vertical (so previous column must be mixed).
     • dp[i][1] = ways with column i "mixed" (part horizontal): can attach a horizontal block of width n or width n−2 to a previous fully vertical boundary.

3. Full solution approach
1) Read m,n. If m·n is odd, print 0 and exit.
2) Ensure n≤m by swapping.
3) Case n=1: print 1.
4) Case n=2: build dp[0…m] with
   dp[0]=1, dp[1]=1, dp[2]=2,
   for i from 3 to m: dp[i] = dp[i−1] + dp[i−3].
   Print dp[m].
5) Case n>2 and odd: build f[0…m], f[0]=1, and for i=1…m do
     if i≥n−1: f[i] += f[i−(n−1)]
     if i≥n+1: f[i] += f[i−(n+1)]
   Print 2·f[m].
6) Case n>2 and even: build dp[0…m][0..1], initialize dp[0][0]=dp[0][1]=1. For i=1…m do:
     dp[i][0] = dp[i−1][1]
     dp[i][1] = (i≥n−2 ? dp[i−(n−2)][0] : 0) + (i≥n ? dp[i−n][0] : 0)
   Print dp[m][0] + dp[m][1].
All arrays store big integers (C++: custom bigint struct; Python: built-ins).

4. C++ implementation with detailed comments
```cpp
#include <bits/stdc++.h>

using namespace std;

template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second;
}

template<typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
    return in >> x.first >> x.second;
}

template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    for(auto& x: a) {
        in >> x;
    }
    return in;
};

template<typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
    for(auto x: a) {
        out << x << ' ';
    }
    return out;
};

// base and base_digits must be consistent
const int base = 1000000000;
const int base_digits = 9;

struct bigint {
    vector<int> z;
    int sign;

    bigint() : sign(1) {}

    bigint(long long v) { *this = v; }

    bigint(const string& s) { read(s); }

    void operator=(const bigint& v) {
        sign = v.sign;
        z = v.z;
    }

    void operator=(long long v) {
        sign = 1;
        if(v < 0) {
            sign = -1, v = -v;
        }
        z.clear();
        for(; v > 0; v = v / base) {
            z.push_back(v % base);
        }
    }

    bigint operator+(const bigint& v) const {
        if(sign == v.sign) {
            bigint res = v;

            for(int i = 0, carry = 0;
                i < (int)max(z.size(), v.z.size()) || carry; ++i) {
                if(i == (int)res.z.size()) {
                    res.z.push_back(0);
                }
                res.z[i] += carry + (i < (int)z.size() ? z[i] : 0);
                carry = res.z[i] >= base;
                if(carry) {
                    res.z[i] -= base;
                }
            }
            return res;
        }
        return *this - (-v);
    }

    bigint operator-(const bigint& v) const {
        if(sign == v.sign) {
            if(abs() >= v.abs()) {
                bigint res = *this;
                for(int i = 0, carry = 0; i < (int)v.z.size() || carry; ++i) {
                    res.z[i] -= carry + (i < (int)v.z.size() ? v.z[i] : 0);
                    carry = res.z[i] < 0;
                    if(carry) {
                        res.z[i] += base;
                    }
                }
                res.trim();
                return res;
            }
            return -(v - *this);
        }
        return *this + (-v);
    }

    void operator*=(int v) {
        if(v < 0) {
            sign = -sign, v = -v;
        }
        for(int i = 0, carry = 0; i < (int)z.size() || carry; ++i) {
            if(i == (int)z.size()) {
                z.push_back(0);
            }
            long long cur = z[i] * (long long)v + carry;
            carry = (int)(cur / base);
            z[i] = (int)(cur % base);
            // asm("divl %%ecx" : "=a"(carry), "=d"(a[i]) : "A"(cur),
            // "c"(base));
        }
        trim();
    }

    bigint operator*(int v) const {
        bigint res = *this;
        res *= v;
        return res;
    }

    friend pair<bigint, bigint> divmod(const bigint& a1, const bigint& b1) {
        int norm = base / (b1.z.back() + 1);
        bigint a = a1.abs() * norm;
        bigint b = b1.abs() * norm;
        bigint q, r;
        q.z.resize(a.z.size());

        for(int i = a.z.size() - 1; i >= 0; i--) {
            r *= base;
            r += a.z[i];
            int s1 = b.z.size() < r.z.size() ? r.z[b.z.size()] : 0;
            int s2 = b.z.size() - 1 < r.z.size() ? r.z[b.z.size() - 1] : 0;
            int d = ((long long)s1 * base + s2) / b.z.back();
            r -= b * d;
            while(r < 0) {
                r += b, --d;
            }
            q.z[i] = d;
        }

        q.sign = a1.sign * b1.sign;
        r.sign = a1.sign;
        q.trim();
        r.trim();
        return make_pair(q, r / norm);
    }

    friend bigint sqrt(const bigint& a1) {
        bigint a = a1;
        while(a.z.empty() || a.z.size() % 2 == 1) {
            a.z.push_back(0);
        }

        int n = a.z.size();

        int firstDigit = (int)sqrt((double)a.z[n - 1] * base + a.z[n - 2]);
        int norm = base / (firstDigit + 1);
        a *= norm;
        a *= norm;
        while(a.z.empty() || a.z.size() % 2 == 1) {
            a.z.push_back(0);
        }

        bigint r = (long long)a.z[n - 1] * base + a.z[n - 2];
        firstDigit = (int)sqrt((double)a.z[n - 1] * base + a.z[n - 2]);
        int q = firstDigit;
        bigint res;

        for(int j = n / 2 - 1; j >= 0; j--) {
            for(;; --q) {
                bigint r1 =
                    (r - (res * 2 * base + q) * q) * base * base +
                    (j > 0 ? (long long)a.z[2 * j - 1] * base + a.z[2 * j - 2]
                           : 0);
                if(r1 >= 0) {
                    r = r1;
                    break;
                }
            }
            res *= base;
            res += q;

            if(j > 0) {
                int d1 =
                    res.z.size() + 2 < r.z.size() ? r.z[res.z.size() + 2] : 0;
                int d2 =
                    res.z.size() + 1 < r.z.size() ? r.z[res.z.size() + 1] : 0;
                int d3 = res.z.size() < r.z.size() ? r.z[res.z.size()] : 0;
                q = ((long long)d1 * base * base + (long long)d2 * base + d3) /
                    (firstDigit * 2);
            }
        }

        res.trim();
        return res / norm;
    }

    bigint operator/(const bigint& v) const { return divmod(*this, v).first; }

    bigint operator%(const bigint& v) const { return divmod(*this, v).second; }

    void operator/=(int v) {
        if(v < 0) {
            sign = -sign, v = -v;
        }
        for(int i = (int)z.size() - 1, rem = 0; i >= 0; --i) {
            long long cur = z[i] + rem * (long long)base;
            z[i] = (int)(cur / v);
            rem = (int)(cur % v);
        }
        trim();
    }

    bigint operator/(int v) const {
        bigint res = *this;
        res /= v;
        return res;
    }

    int operator%(int v) const {
        if(v < 0) {
            v = -v;
        }
        int m = 0;
        for(int i = z.size() - 1; i >= 0; --i) {
            m = (z[i] + m * (long long)base) % v;
        }
        return m * sign;
    }

    void operator+=(const bigint& v) { *this = *this + v; }
    void operator-=(const bigint& v) { *this = *this - v; }
    void operator*=(const bigint& v) { *this = *this * v; }
    void operator/=(const bigint& v) { *this = *this / v; }

    bool operator<(const bigint& v) const {
        if(sign != v.sign) {
            return sign < v.sign;
        }
        if(z.size() != v.z.size()) {
            return z.size() * sign < v.z.size() * v.sign;
        }
        for(int i = z.size() - 1; i >= 0; i--) {
            if(z[i] != v.z[i]) {
                return z[i] * sign < v.z[i] * sign;
            }
        }
        return false;
    }

    bool operator>(const bigint& v) const { return v < *this; }
    bool operator<=(const bigint& v) const { return !(v < *this); }
    bool operator>=(const bigint& v) const { return !(*this < v); }
    bool operator==(const bigint& v) const {
        return !(*this < v) && !(v < *this);
    }
    bool operator!=(const bigint& v) const { return *this < v || v < *this; }

    void trim() {
        while(!z.empty() && z.back() == 0) {
            z.pop_back();
        }
        if(z.empty()) {
            sign = 1;
        }
    }

    bool isZero() const { return z.empty() || (z.size() == 1 && !z[0]); }

    bigint operator-() const {
        bigint res = *this;
        res.sign = -sign;
        return res;
    }

    bigint abs() const {
        bigint res = *this;
        res.sign *= res.sign;
        return res;
    }

    long long longValue() const {
        long long res = 0;
        for(int i = z.size() - 1; i >= 0; i--) {
            res = res * base + z[i];
        }
        return res * sign;
    }

    friend bigint gcd(const bigint& a, const bigint& b) {
        return b.isZero() ? a : gcd(b, a % b);
    }
    friend bigint lcm(const bigint& a, const bigint& b) {
        return a / gcd(a, b) * b;
    }

    void read(const string& s) {
        sign = 1;
        z.clear();
        int pos = 0;
        while(pos < (int)s.size() && (s[pos] == '-' || s[pos] == '+')) {
            if(s[pos] == '-') {
                sign = -sign;
            }
            ++pos;
        }
        for(int i = s.size() - 1; i >= pos; i -= base_digits) {
            int x = 0;
            for(int j = max(pos, i - base_digits + 1); j <= i; j++) {
                x = x * 10 + s[j] - '0';
            }
            z.push_back(x);
        }
        trim();
    }

    friend istream& operator>>(istream& stream, bigint& v) {
        string s;
        stream >> s;
        v.read(s);
        return stream;
    }

    friend ostream& operator<<(ostream& stream, const bigint& v) {
        if(v.sign == -1) {
            stream << '-';
        }
        stream << (v.z.empty() ? 0 : v.z.back());
        for(int i = (int)v.z.size() - 2; i >= 0; --i) {
            stream << setw(base_digits) << setfill('0') << v.z[i];
        }
        return stream;
    }

    static vector<int> convert_base(
        const vector<int>& a, int old_digits, int new_digits
    ) {
        vector<long long> p(max(old_digits, new_digits) + 1);
        p[0] = 1;
        for(int i = 1; i < (int)p.size(); i++) {
            p[i] = p[i - 1] * 10;
        }
        vector<int> res;
        long long cur = 0;
        int cur_digits = 0;
        for(int i = 0; i < (int)a.size(); i++) {
            cur += a[i] * p[cur_digits];
            cur_digits += old_digits;
            while(cur_digits >= new_digits) {
                res.push_back(int(cur % p[new_digits]));
                cur /= p[new_digits];
                cur_digits -= new_digits;
            }
        }
        res.push_back((int)cur);
        while(!res.empty() && res.back() == 0) {
            res.pop_back();
        }
        return res;
    }

    typedef vector<long long> vll;

    static vll karatsubaMultiply(const vll& a, const vll& b) {
        int n = a.size();
        vll res(n + n);
        if(n <= 32) {
            for(int i = 0; i < n; i++) {
                for(int j = 0; j < n; j++) {
                    res[i + j] += a[i] * b[j];
                }
            }
            return res;
        }

        int k = n >> 1;
        vll a1(a.begin(), a.begin() + k);
        vll a2(a.begin() + k, a.end());
        vll b1(b.begin(), b.begin() + k);
        vll b2(b.begin() + k, b.end());

        vll a1b1 = karatsubaMultiply(a1, b1);
        vll a2b2 = karatsubaMultiply(a2, b2);

        for(int i = 0; i < k; i++) {
            a2[i] += a1[i];
        }
        for(int i = 0; i < k; i++) {
            b2[i] += b1[i];
        }

        vll r = karatsubaMultiply(a2, b2);
        for(int i = 0; i < (int)a1b1.size(); i++) {
            r[i] -= a1b1[i];
        }
        for(int i = 0; i < (int)a2b2.size(); i++) {
            r[i] -= a2b2[i];
        }

        for(int i = 0; i < (int)r.size(); i++) {
            res[i + k] += r[i];
        }
        for(int i = 0; i < (int)a1b1.size(); i++) {
            res[i] += a1b1[i];
        }
        for(int i = 0; i < (int)a2b2.size(); i++) {
            res[i + n] += a2b2[i];
        }
        return res;
    }

    bigint operator*(const bigint& v) const {
        vector<int> a6 = convert_base(this->z, base_digits, 6);
        vector<int> b6 = convert_base(v.z, base_digits, 6);
        vll a(a6.begin(), a6.end());
        vll b(b6.begin(), b6.end());
        while(a.size() < b.size()) {
            a.push_back(0);
        }
        while(b.size() < a.size()) {
            b.push_back(0);
        }
        while(a.size() & (a.size() - 1)) {
            a.push_back(0), b.push_back(0);
        }
        vll c = karatsubaMultiply(a, b);
        bigint res;
        res.sign = sign * v.sign;
        for(int i = 0, carry = 0; i < (int)c.size(); i++) {
            long long cur = c[i] + carry;
            res.z.push_back((int)(cur % 1000000));
            carry = (int)(cur / 1000000);
        }
        res.z = convert_base(res.z, 6, base_digits);
        res.trim();
        return res;
    }
};

int n, m;

void read() { cin >> n >> m; }

void solve() {
    // Count cornerless domino tilings of an n x m board (no four dominoes
    // meeting at a corner). Results are huge, so accumulate with bigint.
    // Assume n <= m by swapping.
    //
    // If n * m is odd there is no tiling. If n == 1 there is exactly one.
    //
    // n == 2: dp[i] = ways to fill a 2 x i strip. Either one vertical
    // domino at the end (dp[i - 1]) or two horizontal dominoes capped by a
    // vertical one (dp[i - 3]); three horizontal in a row would create a
    // bad corner. So dp[i] = dp[i - 1] + dp[i - 3].
    //
    // n >= 3: any K x K block has a unique cornerless filling once the
    // top-right tile's orientation is fixed (two mirror configurations).
    // Adjacent equal-orientation pairs cannot be extended a third time.
    //
    // n even: dp[i][0] = first i columns end in a fully vertical column,
    // dp[i][1] = otherwise. dp[i][0] = dp[i - 1][1] (no two adjacent full
    // vertical columns). dp[i][1] = dp[i - n][0] + dp[i - (n - 2)][0], the
    // two block-completion options. Answer is dp[m][0] + dp[m][1].
    //
    // n odd: a fully vertical column is impossible, so a 1D dp suffices.
    // From dp[i] we can reach dp[i - (n - 1)] and dp[i - (n + 1)], each in
    // a unique way up to the corner tile orientation; the global factor of
    // two for the topmost/rightmost orientation gives 2 * dp[m].

    if((int64_t)n * m % 2 == 1) {
        cout << 0 << '\n';
        return;
    }

    if(n > m) {
        swap(n, m);
    }

    if(n == 1) {
        cout << 1 << '\n';
        return;
    }

    if(n == 2) {
        vector<bigint> dp(m + 1, bigint(0));
        dp[0] = 1;
        dp[1] = 1;
        dp[2] = 2;
        for(int i = 3; i <= m; i++) {
            dp[i] = dp[i - 1] + dp[i - 3];
        }
        cout << dp[m] << '\n';
        return;
    }

    if(n % 2 == 1) {
        vector<bigint> dp(m + 1, bigint(0));
        dp[0] = 1;
        for(int i = 1; i <= m; i++) {
            for(int delta: {n - 1, n + 1}) {
                if(i - delta >= 0) {
                    dp[i] += dp[i - delta];
                }
            }
        }
        cout << dp[m] * 2 << '\n';
        return;
    }

    vector<array<bigint, 2>> dp(m + 1, {bigint(0), bigint(0)});
    dp[0][0] = 1;
    dp[0][1] = 1;
    for(int i = 1; i <= m; i++) {
        dp[i][0] = dp[i - 1][1];
        for(int delta: {n - 2, n}) {
            if(i - delta >= 0) {
                dp[i][1] += dp[i - delta][0];
            }
        }
    }

    cout << dp[m][0] + dp[m][1] << '\n';
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    int T = 1;
    // cin >> T;
    for(int test = 1; test <= T; test++) {
        read();
        // cout << "Case #" << test << ": ";
        solve();
    }

    return 0;
}
```

5. Python implementation with detailed comments
```python
# Uses Python's built-in big integers automatically.
import sys

def main():
    data = sys.stdin.read().strip().split()
    m, n = map(int, data)
    # If area is odd, no tiling possible
    if (m * n) & 1:
        print(0)
        return

    # Work with n <= m
    if n > m:
        n, m = m, n

    # Case n = 1: only one horizontal packing
    if n == 1:
        print(1)
        return

    # Case n = 2: dp[i] = dp[i-1] + dp[i-3]
    if n == 2:
        dp = [0] * (m + 1)
        dp[0], dp[1], dp[2] = 1, 1, 2
        for i in range(3, m + 1):
            dp[i] = dp[i - 1] + dp[i - 3]
        print(dp[m])
        return

    # Case n > 2 and odd: f[i] = f[i-(n-1)] + f[i-(n+1)], answer = 2*f[m]
    if n % 2 == 1:
        dp = [0] * (m + 1)
        dp[0] = 1
        for i in range(1, m + 1):
            if i - (n - 1) >= 0:
                dp[i] += dp[i - (n - 1)]
            if i - (n + 1) >= 0:
                dp[i] += dp[i - (n + 1)]
        print(dp[m] * 2)
        return

    # Case n > 2 and even: two-state DP
    # dp[i][0]: ends with vertical column; dp[i][1]: ends mixed
    dp = [[0, 0] for _ in range(m + 1)]
    dp[0][0] = dp[0][1] = 1
    for i in range(1, m + 1):
        # vertical at i => previous must be mixed
        dp[i][0] = dp[i - 1][1]
        # mixed at i => add a block of width n-2 or n from previous vertical
        if i - (n - 2) >= 0:
            dp[i][1] += dp[i - (n - 2)][0]
        if i - n >= 0:
            dp[i][1] += dp[i - n][0]
    # total ways is sum of both ending states
    print(dp[m][0] + dp[m][1])

if __name__ == "__main__":
    main()
```

This completes a step-by-step method to count all cornerless tilings of an m×n rectangle in O(m) time using big integers.
