Given a matrix (H × W) which contains only 1 and 0, find the area of the largest rectangle which only contains 0s.

Constraints

* 1 ≤ H, W ≤ 1,400

Input


H W
c1,1 c1,2 ... c1,W
c2,1 c2,2 ... c2,W
:
cH,1 cH,2 ... cH,W


In the first line, two integers H and W separated by a space character are given. In the following H lines, ci,j, elements of the H × W matrix, are given.

Output

Print the area (the number of 0s) of the largest rectangle.

Example

Input

4 5
0 0 1 0 0
1 0 0 0 0
0 0 0 1 0
0 0 0 1 0


Output

6
#include <bits/stdc++.h>
using namespace std;
int main() {
  int h, w;
  cin >> h >> w;
  int64_t dp[h][w + 1];
  for (int i = 0; i < h; i++) {
    for (int j = 0; j < w; j++) {
      cin >> dp[i][j];
      dp[i][j] = !dp[i][j];
    }
    dp[i][w] = 0;
  }
  for (int i = 1; i < h; i++) {
    for (int j = 0; j < w; j++) {
      if (dp[i][j] == 0) continue;
      dp[i][j] += dp[i - 1][j];
    }
  }
  for (int i = 0; i < h; i++) {
    for (int j = 0; j < w + 1; j++) cerr << dp[i][j] << ' ';
    cerr << endl;
  }
  int64_t ans = 0;
  for (int i = 0; i < h; i++) {
    stack<int64_t> st;
    for (int j = 0; j < w + 1; j++) {
      if (st.empty()) {
        st.emplace(dp[i][j]);
        continue;
      }
      for (int64_t x = 2; !st.empty() && dp[i][j] < st.top(); x++) {
        st.pop();
        ans = max(ans, x * dp[i][j]);
      }
      st.emplace(dp[i][j]);
    }
  }
  cout << ans << endl;
}
