<|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++: boost::multiprecision::cpp_int; Python: built-ins).

4. C++ implementation with detailed comments  
```cpp
#include <bits/stdc++.h>
#include <boost/multiprecision/cpp_int.hpp>
using namespace std;
using bigint = boost::multiprecision::cpp_int;

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

    int m,n;
    cin >> m >> n;
    // If area is odd, no tiling possible
    if ((long long)m * n % 2 == 1) {
        cout << 0 << "\n";
        return 0;
    }
    // Ensure n <= m
    if (n > m) swap(n, m);

    // Case n = 1: only one way (all dominoes along the strip)
    if (n == 1) {
        cout << 1 << "\n";
        return 0;
    }

    // Case n = 2: 1D DP dp[i] = dp[i-1] + dp[i-3]
    if (n == 2) {
        vector<bigint> dp(m+1);
        dp[0] = 1;  // empty strip
        dp[1] = 1;  // one vertical domino only
        dp[2] = 2;  // either two vertical or two horizontals
        for (int i = 3; i <= m; ++i) {
            dp[i] = dp[i-1] + dp[i-3];
        }
        cout << dp[m] << "\n";
        return 0;
    }

    // Case n > 2 and odd: single-state DP with two jump sizes (n-1) and (n+1)
    if (n % 2 == 1) {
        vector<bigint> f(m+1);
        f[0] = 1;
        for (int i = 1; i <= m; ++i) {
            if (i >= n-1) f[i] += f[i - (n-1)];
            if (i >= n+1) f[i] += f[i - (n+1)];
        }
        // Two global orientations
        cout << f[m] * 2 << "\n";
        return 0;
    }

    // Case n > 2 and even: two-state DP
    // dp[i][0] = ways where column i is fully vertical
    // dp[i][1] = ways where column i has horizontal blocks
    vector<array<bigint,2>> dp(m+1);
    dp[0][0] = 1;
    dp[0][1] = 1;  // base: empty prefix counts as both for transitions
    for (int i = 1; i <= m; ++i) {
        // To end with a vertical column at i, previous must be mixed
        dp[i][0] = dp[i-1][1];
        // To end mixed, attach a horizontal block of width (n-2) or n to a vertical boundary
        if (i >= n-2) dp[i][1] += dp[i - (n-2)][0];
        if (i >= n)   dp[i][1] += dp[i - n][0];
    }
    cout << dp[m][0] + dp[m][1] << "\n";
    return 0;
}
```

5. Python implementation with detailed comments  
```python
import sys
def main():
    data = sys.stdin.read().split()
    m, n = map(int, data)
    # If area is odd, no tiling
    if (m * n) & 1:
        print(0)
        return

    # Ensure 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: single-state DP with jumps (n-1),(n+1)
    if n % 2 == 1:
        f = [0] * (m + 1)
        f[0] = 1
        for i in range(1, m + 1):
            if i >= n-1:
                f[i] += f[i - (n-1)]
            if i >= n+1:
                f[i] += f[i - (n+1)]
        # Two orientations
        print(f[m] * 2)
        return

    # Case n > 2 and even: two-state DP
    # dp[i][0]: column i fully vertical; dp[i][1]: ends with horizontal blocks
    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 => prev must be mixed
        dp[i][0] = dp[i-1][1]
        # mixed at i => attach horizontal block width n-2 or n to prev vertical
        if i >= n-2:
            dp[i][1] += dp[i - (n-2)][0]
        if i >= n:
            dp[i][1] += dp[i - n][0]
    # total ways is sum of both states at column m
    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.