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 main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  int N, M;
  cin >> N >> M;
  vector<vector<int> > A(N, vector<int>(M));
  for (int i = 0; i < N * M; i++) cin >> A[i / M][i % M];
  int ans = 1;
  vector<int> occL(160000 + 47, -1);
  vector<vector<vector<int> > > lastR(
      M, vector<vector<int> >(M, vector<int>(N, -1)));
  for (int i = 0; i < M; i++)
    for (int j = i; j < M; j++) {
      for (int k = 0; k < N; k++) {
        if (i != j && A[k][i] == A[k][j])
          lastR[i][j][k] = k;
        else
          lastR[i][j][k] = max(occL[A[k][i]], occL[A[k][j]]);
        occL[A[k][i]] = occL[A[k][j]] = k;
      }
      for (int k = 0; k < N; k++) occL[A[k][i]] = occL[A[k][j]] = -1;
    }
  for (int l = 2; l <= M; l++)
    for (int i = 0; i <= M - l; i++)
      for (int k = 0; k < N; k++)
        lastR[i][i + l - 1][k] =
            max(lastR[i][i + l - 1][k],
                max(lastR[i + 1][i + l - 1][k], lastR[i][i + l - 2][k]));
  for (int i = 0; i < M; i++)
    for (int j = i; j < M; j++) {
      int last = -1;
      for (int k = 0; k < N; k++) {
        last = max(last, lastR[i][j][k]);
        ans = max(ans, (j - i + 1) * (k - last));
      }
    }
  cout << ans << "\n";
  return 0;
}
