1. Abridged Problem Statement
Given an N×M grid (1≤M≤5, N up to 10^100) and two colors (black/white), count the number of ways to color every cell so that no 2×2 sub‐square is monochrome. Output the count modulo P (1≤P≤10^4).

2. Detailed Editorial

Overview
We scan the grid row by row. Each row’s coloring can be represented by an M‐bit mask (0=white, 1=black). We define a transition rule between two consecutive rows: they form no 2×2 monochrome square. Let S=2^M.

DP Formulation
Let dp[i][mask] = number of valid colorings of rows 1..i ending with row i colored as mask. Then
 dp[1][mask] = 1 for all mask in [0, S).
 dp[i][cur] = ∑_{prev=0 to S−1} dp[i−1][prev] * valid(prev, cur).

Here valid(prev,cur)=1 if for every k from 1 to M−1 the four bits
 prev at k−1, prev at k, cur at k−1, cur at k
are not all 0 and not all 1.

Matrix Exponentiation
We rewrite the recurrence as a vector–matrix product. Let V_i be the S×1 column vector of dp[i][·]. Define the S×S transition matrix T by
 T[prev][cur] = valid(prev,cur).
Then V_i = T · V_{i−1}, so V_N = T^(N−1) · V_1. Since V_1 is all ones, the total count is just the sum of all entries of T^(N−1). Because N can be up to 10^100, we must exponentiate T to a huge exponent modulo P.

Fast exponentiation with a big exponent
N is represented with a big‐integer type. We form the exponent N−1 and run binary exponentiation, reading its bits via the big‐integer `% 2` and `/= 2` operations:
- While exponent > 0:
  – If exponent is odd, multiply result ← result×T mod P
  – T ← T×T mod P
  – exponent ← exponent / 2

Time Complexity
- Building T: O(S²·M), with S≤32, M≤5
- Each matrix multiply: O(S³) ≤ 32³≈32768 operations
- Exponentiation steps: O((log N)·S³). log N≈330 bits for 10^100. Fast enough under 0.5s with P small.

3. C++ Solution
```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;
    }
};

bigint n;
int m, p;

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

vector<vector<int>> mat_mul(
    const vector<vector<int>>& a, const vector<vector<int>>& b
) {
    int sz = a.size();
    vector<vector<int>> c(sz, vector<int>(sz, 0));
    for(int i = 0; i < sz; i++) {
        for(int j = 0; j < sz; j++) {
            int64_t acc = 0;
            for(int k = 0; k < sz; k++) {
                acc += (int64_t)a[i][k] * b[k][j];
            }
            c[i][j] = acc % p;
        }
    }

    return c;
}

void solve() {
    // - A valid pattern has no monochromatic 2x2 square. Encode each row as a
    //   bitmask over M columns; a pair of consecutive rows is compatible iff for
    //   every adjacent column pair the four cells are not all equal, i.e. not all
    //   ones and not all zeros across the two rows.
    //
    // - Build the (2^M) x (2^M) transition matrix T where T[cur][next] = 1 if
    //   rows cur and next can be stacked. The number of valid N-row patterns is
    //   the sum of all entries of T^(N-1) times the all-ones start vector, all
    //   taken modulo P.
    //
    // - N is up to 10^100, so the exponent N-1 is handled as a big integer and
    //   binary exponentiation reads its bits via the modulo-2 / divide-by-2
    //   operations on the big integer type.

    int states = 1 << m;
    vector<vector<int>> trans(states, vector<int>(states, 0));
    for(int cur = 0; cur < states; cur++) {
        for(int nxt = 0; nxt < states; nxt++) {
            bool valid = true;
            for(int i = 1; i < m; i++) {
                bool all_one = ((cur >> (i - 1)) & 1) && ((cur >> i) & 1) &&
                               ((nxt >> (i - 1)) & 1) && ((nxt >> i) & 1);
                bool all_zero = !((cur >> (i - 1)) & 1) && !((cur >> i) & 1) &&
                                !((nxt >> (i - 1)) & 1) && !((nxt >> i) & 1);
                if(all_one || all_zero) {
                    valid = false;
                    break;
                }
            }

            if(valid) {
                trans[cur][nxt] = 1;
            }
        }
    }

    vector<vector<int>> result(states, vector<int>(states, 0));
    for(int i = 0; i < states; i++) {
        result[i][i] = 1 % p;
    }

    bigint e = n - bigint(1);
    while(e > bigint(0)) {
        if(e % 2) {
            result = mat_mul(result, trans);
        }

        trans = mat_mul(trans, trans);
        e /= 2;
    }

    int64_t ans = 0;
    for(int i = 0; i < states; i++) {
        for(int j = 0; j < states; j++) {
            ans += result[i][j];
        }
    }

    cout << ans % p << '\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;
}
```

4. Python Solution with Detailed Comments
```python
def matrix_multiply(A, B, mod):
    # Multiply two square matrices A and B under modulo mod
    n = len(A)
    C = [[0] * n for _ in range(n)]
    for i in range(n):
        for k in range(n):
            if A[i][k]:
                for j in range(n):
                    C[i][j] = (C[i][j] + A[i][k] * B[k][j]) % mod
    return C

def matrix_vector_multiply(A, v, mod):
    # Multiply matrix A (n×n) by vector v (length n) under modulo mod
    n = len(A)
    result = [0] * n
    for i in range(n):
        s = 0
        for j in range(n):
            s += A[i][j] * v[j]
        result[i] = s % mod
    return result

def matrix_power(matrix, power, mod):
    # Fast exponentiation of square matrix to integer power under mod
    n = len(matrix)
    # Initialize result as identity matrix
    result = [[1 if i == j else 0 for j in range(n)] for i in range(n)]
    base = matrix
    while power > 0:
        if power & 1:
            result = matrix_multiply(result, base, mod)
        base = matrix_multiply(base, base, mod)
        power >>= 1
    return result

def count_nice_patterns(N, M, P):
    S = 1 << M  # number of bitmasks for a row
    # Build transition matrix: from current row mask to next row mask
    T = [[0] * S for _ in range(S)]
    for cur in range(S):
        for nxt in range(S):
            valid = True
            # Check every adjacent column pair for 2×2 mono block
            for i in range(1, M):
                a = (cur >> (i-1)) & 1
                b = (cur >> i) & 1
                c = (nxt >> (i-1)) & 1
                d = (nxt >> i) & 1
                # If all four bits are identical, it's invalid
                if (a + b + c + d) in (0, 4):
                    valid = False
                    break
            if valid:
                T[cur][nxt] = 1

    # Initial state: row 1 can be any mask
    start = [1] * S
    # Compute T^(N-1) under mod P
    Texp = matrix_power(T, N - 1, P)
    # Multiply by initial vector to get counts for row N
    final = matrix_vector_multiply(Texp, start, P)
    # Sum over all end‐masks
    return sum(final) % P

# Read input and output answer
N, M, P = map(int, input().split())
print(count_nice_patterns(N, M, P))
```

5. Compressed Editorial
We map each row to a bitmask of length M. Define an S×S transition matrix T (S=2^M) with T[a][b]=1 if rows a and b do not form any 2×2 monochrome block. The total number is the sum of entries in T^(N−1)·[1…1]^T, all computed modulo P. Since N is up to 10^100, we keep N as a big integer and exponentiate via binary exponentiation, reading the exponent's bits with the big-integer mod-2 / divide-by-2 operations. The cost is O(S³·log N), feasible for S≤32.
