#include <iostream>
#include <cstdio>
#include <vector>
#include <array>
#include <algorithm>
#include <numeric>
#include <set>
#include <queue>
#include <stack>
#include <bitset>
#include <map>
#include <cstring>

using i64 = long long;

int n, m, pre1[1510][1510], pre2[1510][1510];

bool check(int x, int y, int xx, int yy)
{
    if (x == xx || y == yy)
    {
        return 1;
    }

    bool tmp = 0;
    for (int i = x + 1; i < xx; ++i)
    {
        int t = pre1[i][yy] - pre1[i][y];
        if (t == yy - y)
        {
            return (check(x, y, i, yy) && check(i, y, xx, yy));
        }
        else
        {
            tmp = (t > 0);
        }
    }
    for (int i = y + 1; i < yy; ++i)
    {
        int t = pre2[xx][i] - pre2[x][i];
        if (t == xx - x)
        {
            return (check(x, y, xx, i) && check(x, i, xx, yy));
        }
        else
        {
            tmp = (t > 0);
        }
    }

    return !tmp; // 检查是否有剩余无法拼接
}

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

    std::cin >> n >> m;
    for (int i = 1; i < n; ++i)
    {
        for (int j = 1; j <= m; ++j)
        {
            char c;
            std::cin >> c;
            pre1[i][j] += pre1[i][j - 1] + c - '0';
        }
    }

    for (int i = 1; i <= n; ++i)
    {
        for (int j = 1; j < m; ++j)
        {
            char c;
            std::cin >> c;
            pre2[i][j] += pre2[i - 1][j] + c - '0';
        }
    }

    if (check(0, 0, n, m))
    {
        std::cout << "YES\n";
    }
    else
    {
        std::cout << "NO\n";
    }
    return 0;
}
// 复杂度:拼接的顺序不重要,只要存在能横跨两侧数量的墙就可以分割
// T(n)=O(n+m)+O(n+m)+.....+O(n+m)=O((n+m)^2)
//即复杂度只与切割次数有关,最多有n^2条边
//不应该区分块,用墙的数量判断