#include <bits/stdc++.h>
using namespace std;
template <class T>
inline T min(T a, T b, T c) {
  return min(min(a, b), c);
}
template <class T>
inline T max(T a, T b, T c) {
  return max(max(a, b), c);
}
template <class T>
inline void checkmin(T &a, T b) {
  if (b < a) a = b;
}
template <class T>
inline void checkmax(T &a, T b) {
  if (b > a) a = b;
}
template <class T>
inline T sq(T x) {
  return x * x;
}
template <class T>
inline int countbit(T n) {
  return (n == 0) ? 0 : 1 + countbit(n & (n - 1));
}
template <class T>
inline T gcd(T a, T b) {
  return b == 0 ? a : gcd(b, a % b);
}
template <class T>
inline T lcm(T a, T b) {
  return a / gcd(a, b) * b;
}
inline int input() {
  int x = 0;
  char ch = getchar();
  bool flag = true;
  for (; ch < '0' || ch > '9'; ch = getchar())
    if (ch == '-') flag = false;
  for (; ch >= '0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - '0';
  return flag ? x : -x;
}
void output(int x) {
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  int len = 0, dig[20];
  while (x) {
    dig[len++] = x % 10;
    x /= 10;
  }
  if (len == 0) dig[len++] = 0;
  while (len--) putchar(dig[len] + 48);
  putchar('\n');
}
const int N = 405;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
int a[N][N];
int f[N][N];
int last[N * N][N];
int main() {
  int n, m;
  n = input(), m = input();
  for (int i = (0); i < (n); ++i)
    for (int j = (0); j < (m); ++j) a[i][j] = input();
  int ans = 0;
  memset(f, -1, sizeof(f)), memset(last, -1, sizeof(last));
  for (int i = (0); i < (n); ++i) {
    for (int j = (0); j < (m); ++j) {
      for (int k = (0); k < (m); ++k) {
        int l = j, r = k;
        if (l > r) swap(l, r);
        f[l][r] = max(f[l][r], last[a[i][j]][k]);
      }
      last[a[i][j]][j] = i;
    }
    for (int len = (1); len <= (m - 1); ++len) {
      for (int l = (0); l < (m - len); ++l) {
        int r = l + len;
        f[l][r] = max(f[l][r], f[l + 1][r], f[l][r - 1]);
      }
    }
    for (int l = (0); l < (m); ++l) {
      for (int r = (l); r <= (m - 1); ++r) {
        ans = max(ans, (i - f[l][r]) * (r - l + 1));
      }
    }
  }
  printf("%d\n", ans);
  return 0;
}
