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 W, H;
  cin >> H >> W;
  int dp[H][W];
  for (int i = 0; i < H; i++) {
    for (int j = 0; j < W; j++) {
      int tile;
      if (cin >> tile, tile == 0) {
        dp[i][j] = (j ? dp[i][j - 1] : 0) + 1;
      } else {
        dp[i][j] = 0;
      }
    }
  }
  int ans = 0;
  stack<pair<int, int>, vector<pair<int, int>>> S;
  for (int j = 0; j < W; j++) {
    for (int i = 0; i < H; i++) {
      if (S.empty()) {
        S.emplace(dp[i][j], i);
        continue;
      }
      pair<int, int> p = S.top();
      if (p.first == dp[i][j]) {
        continue;
      }
      if (p.first < dp[i][j]) {
        S.emplace(dp[i][j], i);
        continue;
      }
      while (!S.empty()) {
        ans = max(ans, p.first * (i - p.second));
        S.pop();
        if (S.empty()) {
          S.emplace(dp[i][j], p.second);
          break;
        }
        if (S.top().first >= dp[i][j]) {
          p = S.top();
          S.pop();
          continue;
        } else {
          S.emplace(dp[i][j], p.second);
          break;
        }
      }
    }
    while (!S.empty()) {
      pair<int, int> p = S.top();
      S.pop();
      ans = max(ans, p.first * (H - p.second));
    }
  }
  cout << ans << endl;
}
