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;
const int N = 401;
const int M = 160001;
const int K = 1e6 + 1;
short pos[M][N], max1[N][N];
int ar[N][N], val[K], sz = 0;
vector<int> lis;
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, i, j, k, l, m, cnt = 0, ans = 0;
  cin >> n >> m;
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= m; j++) {
      cin >> ar[i][j];
      lis.push_back(ar[i][j]);
    }
  }
  sort(lis.begin(), lis.end());
  i = j = 0;
  while (i < lis.size()) {
    while (j < lis.size() && lis[i] == lis[j]) {
      j++;
    }
    sz++;
    val[lis[i]] = sz;
    i = j;
  }
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= m; j++) {
      ar[i][j] = val[ar[i][j]];
    }
  }
  for (i = 1; i <= n; i++) {
    for (l = 1; l <= m; l++) {
      for (j = 1; j + l - 1 <= m; j++) {
        k = j + l - 1;
        if (j == k) {
          max1[j][k] = max(max1[j][k], pos[ar[i][j]][j]);
        } else {
          if (ar[i][j] == ar[i][k]) {
            max1[j][k] = i;
          } else {
            max1[j][k] =
                max(max1[j][k], max(max(max1[j + 1][k], pos[ar[i][k]][j]),
                                    max(max1[j][k - 1], pos[ar[i][j]][k])));
          }
        }
        ans = max(ans, l * (i - max1[j][k]));
      }
    }
    for (j = 1; j <= m; j++) {
      pos[ar[i][j]][j] = i;
    }
  }
  cout << ans;
}
