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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int rnd(int l, int r) { return l + rng() % (r - l + 1); }
using ii = pair<int, int>;
using ll = long long;
using ull = unsigned long long;
const int N = 405;
int n, m, a[N][N], f[N][N], d[N][N];
int p[N][N * N];
int32_t main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  if (fopen("407D"
            ".inp",
            "r")) {
    freopen(
        "407D"
        ".inp",
        "r", stdin);
    freopen(
        "407D"
        ".out",
        "w", stdout);
  }
  cin >> n >> m;
  for (int i = 1, _c = n; i <= _c; ++i)
    for (int j = 1, _c = m; j <= _c; ++j) cin >> a[i][j];
  int ans = 0;
  for (int j = 1, _c = m; j <= _c; ++j) {
    for (int l = n, _c = 1; l >= _c; --l) {
      for (int r = l, _c = n; r <= _c; ++r) {
        f[l][r] = max({p[r][a[l][j]], p[l][a[r][j]], f[l + 1][r], f[l][r - 1]});
        if (l != r && a[l][j] == a[r][j]) f[l][r] = j;
        d[l][r] = max(d[l][r], f[l][r]);
        ans = max(ans, (r - l + 1) * (j - d[l][r]));
      }
    }
    for (int i = 1, _c = n; i <= _c; ++i) p[i][a[i][j]] = j;
  }
  cout << ans << '\n';
}
