#include <bits/stdc++.h>

using namespace std;
const int INF = 0x3f3f3f3f;

const int N = 110;
int a[N][N];
int n, m;
int f[N][N];

// dfs解法
/*
2 3
1 3 5
2 4 6

答案：
13
*/

int dfs(int x, int y) {
    if (f[x][y]) return f[x][y];
    if (x > n || y > m) return INF;
    if (x == n && y == m) return a[n][m];
    return f[x][y] = a[x][y] + min(dfs(x + 1, y), dfs(x, y + 1));
}

int main() {
    cin >> n >> m;
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= m; j++)
            cin >> a[i][j];

    printf("%d\n", dfs(1, 1));
    return 0;
}