/*input
2 3
2 7 5
1 9 5
*/
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double ld;
typedef complex<ld> cd;
typedef pair<int, int> pi;
typedef pair<ll, ll> pl;
typedef pair<ld, ld> pd;
typedef vector<int> vi;
typedef vector<vi> vii;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<vl> vll;
typedef vector<pi> vpi;
typedef vector<vpi> vpii;
typedef vector<pl> vpl;
typedef vector<cd> vcd;
typedef vector<pd> vpd;
typedef vector<bool> vb;
typedef vector<vb> vbb;
typedef std::string str;
typedef std::vector<str> vs;
#define x first
#define y second
#define debug(...) cout<<"["<<#__VA_ARGS__<<": "<<__VA_ARGS__<<"]\n"
const int MOD = 1000000007;
const ll INF = std::numeric_limits<ll>::max();
const int MX = 100101;
const ld PI = 3.14159265358979323846264338327950288419716939937510582097494L;
template<typename T>
pair<T, T> operator+(const pair<T, T> &a, const pair<T, T> &b) { return pair<T, T>(a.x + b.x, a.y + b.y); }
template<typename T>
pair<T, T> operator-(const pair<T, T> &a, const pair<T, T> &b) { return pair<T, T>(a.x - b.x, a.y - b.y); }
template<typename T>
T operator*(const pair<T, T> &a, const pair<T, T> &b) { return (a.x * b.x + a.y * b.y); }
template<typename T>
T operator^(const pair<T, T> &a, const pair<T, T> &b) { return (a.x * b.y - a.y * b.x); }
template<typename T>
void print(vector<T> vec, string name = "") {
cout << name;
for (auto u : vec)
cout << u << ' ';
cout << '\n';
}
ll dp[50][50][50][50];
int main() {
ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
int N, M;
cin >> N >> M;
vii sk(N, vi(M));
vii sm(N, vi(M, 0));
for (int i = 0; i < N; ++i)
{
for (int j = 0; j < M; ++j)
{
cin >> sk[i][j];
sm[i][j] = sk[i][j];
if (i) sm[i][j] += sm[i - 1][j];
if (j) sm[i][j] += sm[i][j - 1];
if (i and j) sm[i][j] -= sm[i - 1][j - 1];
}
}
auto sum = [&](int ln, int lm, int hn, int hm) {
return sm[hn][hm]
- (lm ? sm[hn][lm - 1] : 0)
- (ln ? sm[ln - 1][hm] : 0)
+ (lm and ln ? sm[ln - 1][lm - 1] : 0);
};
for (int i = 0; i < N; ++i)
{
for (int j = 0; j < M; ++j)
{
for (int n = 0; n + i < N; ++n)
{
for (int m = 0; m + j < M; ++m)
{
if (!i and !j) {
dp[n][m][n + i][m + j] = 0;
continue;
}
dp[n][m][n + i][m + j] = (ll)1e16;
for (int ki = 0; ki < i; ++ki)
{
ll nauja = dp[n][m][n + ki][m + j]
+ dp[n + ki + 1][m][n + i][m + j];
dp[n][m][n + i][m + j] = min(dp[n][m][n + i][m + j], nauja);
}
for (int kj = 0; kj < j; ++kj)
{
ll nauja = dp[n][m][n + i][m + kj]
+ dp[n][m + kj + 1][n + i][m + j];
dp[n][m][n + i][m + j] = min(dp[n][m][n + i][m + j], nauja);
}
dp[n][m][n + i][m + j] += sum(n, m, n + i, m + j);
}
}
}
}
printf("%lld\n", dp[0][0][N - 1][M - 1]);
}
/* Look for:
* special cases (n=1?)
* overflow (ll vs int?)
* the exact constraints (multiple sets are too slow for n=10^6 :( )
* array bounds
*/
