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;
short f[402][402][402];
int cas[160050];
int pre[160050];
int a[402][402];
int main() {
  int n, m, ri = 0, i, j, k, i1, i2;
  scanf("%d%d", &n, &m);
  for (i = 1; i <= n; i++)
    for (j = 1; j <= m; j++) scanf("%d", &a[i][j]);
  memset(cas, 0, sizeof(cas));
  memset(pre, 0, sizeof(pre));
  for (i1 = 1; i1 <= n; i1++)
    for (i2 = i1; i2 <= n; i2++)
      for (j = 1; j <= m; j++) f[i1][i2][j] = m;
  for (i = 1; i <= n; i++) {
    ri++;
    for (j = m; j > 0; j--) {
      if (cas[a[i][j]] == ri) {
        f[i][i][j] = pre[a[i][j]] - 1;
      }
      cas[a[i][j]] = ri;
      pre[a[i][j]] = j;
    }
  }
  for (i1 = 1; i1 <= n; i1++) {
    for (i2 = i1 + 1; i2 <= n; i2++) {
      ri++;
      for (j = m; j > 0; j--) {
        if (cas[a[i1][j]] == ri && f[i1][i2][j] > pre[a[i1][j]] - 1) {
          f[i1][i2][j] = pre[a[i1][j]] - 1;
        }
        cas[a[i1][j]] = ri;
        pre[a[i1][j]] = j;
        if (cas[a[i2][j]] == ri && f[i1][i2][j] > pre[a[i2][j]] - 1) {
          f[i1][i2][j] = pre[a[i2][j]] - 1;
        }
        cas[a[i2][j]] = ri;
        pre[a[i2][j]] = j;
      }
    }
  }
  int ans = 0;
  for (i1 = n; i1 > 0; i1--) {
    for (i2 = i1; i2 <= n; i2++) {
      for (j = m; j > 0; j--) {
        int now = f[i1][i2][j];
        int area = (i2 - i1 + 1) * (now - j + 1);
        if (area > ans) {
          ans = area;
        }
        if (i1 > 1 && now < f[i1 - 1][i2][j]) f[i1 - 1][i2][j] = now;
        if (i2 < n && now < f[i1][i2 + 1][j]) f[i1][i2 + 1][j] = now;
        if (j > 1 && now < f[i1][i2][j - 1]) f[i1][i2][j - 1] = now;
      }
    }
  }
  printf("%d\n", ans);
}
