You are given matrix a of size n × m, its elements are integers. We will assume that the rows of the matrix are numbered from top to bottom from 1 to n, the columns are numbered from left to right from 1 to m. We will denote the element on the intersecting of the i-th row and the j-th column as aij.

We'll call submatrix i1, j1, i2, j2 (1 ≤ i1 ≤ i2 ≤ n; 1 ≤ j1 ≤ j2 ≤ m) such elements aij of the given matrix that i1 ≤ i ≤ i2 AND j1 ≤ j ≤ j2. We'll call the area of the submatrix number (i2 - i1 + 1)·(j2 - j1 + 1). We'll call a submatrix inhomogeneous, if all its elements are distinct.

Find the largest (in area) inhomogenous submatrix of the given matrix.

Input

The first line contains two integers n, m (1 ≤ n, m ≤ 400) — the number of rows and columns of the matrix, correspondingly.

Each of the next n lines contains m integers aij (1 ≤ aij ≤ 160000) — the elements of the matrix.

Output

Print a single integer — the area of the optimal inhomogenous submatrix.

Examples

Input

3 3
1 3 1
4 5 6
2 6 1


Output

6


Input

3 4
5 2 3 1
3 3 5 3
4 4 4 5


Output

4


Input

2 6
1 2 3 4 5 6
8 6 7 8 9 1


Output

8
#include <bits/stdc++.h>
using namespace std;
int n, m, a[401][401];
short int dp[401][401][401], u[401][401][401], v[160001];
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++) scanf("%d", &a[i][j]);
  memset(dp, 0, sizeof(dp));
  memset(u, 0, sizeof(u));
  memset(v, 0, sizeof(v));
  for (int i = 1; i <= m; i++) {
    int l = n + 1;
    for (int j = n; j >= 1; j--) {
      while (!v[a[l - 1][i]] && l >= 2) {
        v[a[l - 1][i]] = 1;
        l--;
      }
      u[i][i][j] = l;
      v[a[j][i]] = 0;
    }
  }
  for (int i = 1; i <= m; i++)
    for (int j = i + 1; j <= m; j++) {
      int l = n + 1;
      for (int k = n; k >= 1; k--) {
        l = min(l, k + 1);
        while (!v[a[l - 1][i]] && !v[a[l - 1][j]] && l >= 2) {
          if (a[l - 1][i] == a[l - 1][j]) break;
          v[a[l - 1][i]] = v[a[l - 1][j]] = 1;
          l--;
        }
        u[i][j][k] = l;
        v[a[k][i]] = v[a[k][j]] = 0;
      }
    }
  for (int i = 1; i <= m; i++)
    for (int j = 1; j <= n; j++) dp[i][i][j] = u[i][i][j];
  for (int l = 2; l <= m; l++)
    for (int i = 1; i + l - 1 <= m; i++) {
      int j = i + l - 1;
      for (int k = 1; k <= n; k++)
        dp[i][j][k] = (short int)max(
            (short int)((short int)max((short int)(dp[i + 1][j][k]),
                                       (short int)(dp[i][j - 1][k]))),
            (short int)(u[i][j][k]));
    }
  int ans = 0;
  for (int i = 1; i <= m; i++)
    for (int j = i; j <= m; j++)
      for (int k = 1; k <= n; k++)
        ans = max(ans, int(k - dp[i][j][k] + 1) * int(j - i + 1));
  printf("%d\n", ans);
  return 0;
}
