/***************************************
      zzblack                         **
      2015-11-21                      **
      Orz                             **
****************************************/
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <cmath>
#include <string>
#include <cctype>
#include <cstdio>
#include <vector>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <sstream>
#include <iostream>
#include <algorithm>
using namespace std;
#define ls id<<1,l,mid
#define rs id<<1|1,mid+1,r
#define OFF(x) memset(x,-1,sizeof x)
#define CLR(x) memset(x,0,sizeof x)
#define MEM(x) memset(x,0x3f,sizeof x)
typedef long long ll ;
typedef pair<int,int> pii ;
const int maxn = 1e5 + 50 ;
const double eps = 1e-10;
const int max_index = 62;
const int inf = 0x3f3f3f3f ;
const int MOD = 1e9+7 ;

inline int read(){
    char c = getchar();
    while (!isdigit(c)) c = getchar();
    int x = 0;
    while (isdigit(c)) {
        x = x * 10 + c - '0';
        c = getchar();
    }
    return x;
}

int n, m;
int num[1100][1100], dp[1100][1100];

int main () {
#ifdef LOCAL
    freopen("C:\\Users\\Administrator\\Desktop\\in.txt","r",stdin);
//      freopen("C:\\Users\\Administrator\\Desktop\\out.txt","w",stdout);
#endif
    while (~scanf("%d%d", &n, &m)) {
//        printf("%d %d\n", n, m);
        MEM(dp);
        for (int i = 1; i <= n; i++) dp[i][0] = 0;
        for (int j = 1; j <= m; j++) dp[0][j] = 0;
        dp[0][0] = 0;
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= m; j++) {
                scanf("%d", &num[i][j]);
                if ((i + j) % 2 == 0) continue;
                if (i > 2 || j == 1) dp[i][j] = min(dp[i][j], dp[i - 2][j] + num[i][j] * num[i - 1][j]);
                if (j > 2 || i == 1) dp[i][j] = min(dp[i][j], dp[i][j - 2] + num[i][j] * num[i][j - 1]);
                if (i > 1 && j > 1)
                    dp[i][j] = min(dp[i][j], dp[i - 1][j - 1] + min(num[i][j] * num[i - 1][j], num[i][j] * num[i][j - 1]));
            }
        }
        cout << dp[n][m] << "\n";
    }
    return 0;
}