Largest Submatrix 3

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 a ij.

We’ll call submatrix i 1, j 1, i 2, j 2 (1 ≤ i 1 ≤ i 2 ≤ n; 1 ≤ j 1 ≤ j 2 ≤ m) such elements a ij of the given matrix that i 1 ≤ i ≤ i 2 AND j 1 ≤ j ≤ j 2. We’ll call the area of the submatrix number (i 2 - i 1 + 1)·(j 2 - j 1 + 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 nm (1 ≤ n, m ≤ 400) — the number of rows and columns of the matrix, correspondingly.

Each of the next n lines contains m integers a ij (1 ≤ a ij ≤ 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

Solution:

#include <iostream>
#include <iomanip>
#include <cstdio>
#include <set>
#include <vector>
#include <map>
#include <cmath>
#include <algorithm>
#include <memory.h>
#include <string>
#include <cstring>
#include <sstream>
#include <cstdlib>
#include <ctime>
#include <cassert>

using namespace std;

const int N = 444;
const int M = N * N;

short f[N][N][N];

inline void add(int xa, int ya, int xb, int yb) {
yb--;
if (yb < f[xa][xb][ya]) {
    f[xa][xb][ya] = yb;
  }
}

int last[M], was[M];
int a[N][N];

int main() {
  int r, c;
  scanf("%d %d", &r, &c);
  for (int i = 1; i <= r; i++)
    for (int j = 1; j <= c; j++) scanf("%d", a[i] + j);
  for (int i1 = 1; i1 <= r; i1++)
    for (int i2 = i1; i2 <= r; i2++)
      for (int j = 1; j <= c; j++) f[i1][i2][j] = c;
  for (int i = 0; i < M; i++) {
    last[i] = 0;
    was[i] = 0;
  }
  int it = 0;
  for (int i = 1; i <= r; i++) {
    it++;
    for (int j = c; j >= 1; j--) {
if (was[a[i][j]] == it) {
add(i, j, i, last[a[i][j]]);
}
last[a[i][j]] = j;
was[a[i][j]] = it;
}
}
for (int i1 = 1; i1 <= r; i1++) {
    for (int i2 = i1 + 1; i2 <= r; i2++) {
      it++;
      for (int j = c; j >= 1; j--) {
if (was[a[i1][j]] == it) {
add(i1, j, i2, last[a[i1][j]]);
}
last[a[i1][j]] = j;
was[a[i1][j]] = it;
if (was[a[i2][j]] == it) {
add(i1, j, i2, last[a[i2][j]]);
}
last[a[i2][j]] = j;
was[a[i2][j]] = it;
}
}
}
int ans = 0;
for (int i1 = r; i1 >= 1; i1--)
for (int i2 = i1; i2 <= r; i2++)
      for (int j = c; j >= 1; j--) {
int ft = f[i1][i2][j];
int area = (i2 - i1 + 1) * (ft - j + 1);
if (area > ans) {
ans = area;
}
if (i1 > 1 && ft < f[i1 - 1][i2][j]) {
          f[i1 - 1][i2][j] = ft;
        }
        if (i2 < r && ft < f[i1][i2 + 1][j]) {
          f[i1][i2 + 1][j] = ft;
        }
        if (j > 1 && ft < f[i1][i2][j - 1]) {
          f[i1][i2][j - 1] = ft;
        }
      }
  printf("%d\n", ans);
  return 0;
}