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>
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
using namespace std;
int n, m, a[410][410], ans;
int last[1000010];
int L[410][410], R[410][410], limit[410];
int main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= m; ++j) cin >> a[i][j];
  for (int up = n; up >= 1; --up) {
    for (int i = 1; i <= m; ++i) {
      if (last[a[up][i]])
        L[up][i] = last[a[up][i]];
      else
        L[up][i] = 0;
      last[a[up][i]] = i;
      for (int down = up + 1; down <= n; ++down)
        if (last[a[down][i]] && L[down][i] < last[a[down][i]])
          L[down][i] = last[a[down][i]];
    }
    for (int i = 1; i <= m; ++i) last[a[up][i]] = 0;
    for (int i = m; i >= 1; --i) {
      if (last[a[up][i]])
        R[up][i] = last[a[up][i]];
      else
        R[up][i] = m + 1;
      last[a[up][i]] = i;
      for (int down = up + 1; down <= n; ++down)
        if (last[a[down][i]] && R[down][i] > last[a[down][i]])
          R[down][i] = last[a[down][i]];
    }
    for (int i = 1; i <= m; ++i) last[a[up][i]] = 0;
    for (int i = 1; i <= m + 1; ++i) limit[i] = m + 1;
    for (int down = up; down <= n; ++down) {
      for (int i = m; i >= 1; --i) {
        if (limit[i] > limit[i + 1]) limit[i] = limit[i + 1];
        if (limit[i] > R[down][i]) limit[i] = R[down][i];
        if (limit[L[down][i]] > i) limit[L[down][i]] = i;
        if ((limit[i] - i) * (down - up + 1) > ans)
          ans = (limit[i] - i) * (down - up + 1);
      }
    }
  }
  cout << ans;
}
