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;
const int INF = 1 << 29;
const double EPS = 1e-9;
const int MOD = 100000007;
const int dx[] = {1, 0, -1, 0}, dy[] = {0, -1, 0, 1};
const int MAX_HW = 1500;
int H, W;
int board[MAX_HW][MAX_HW];
int dp[MAX_HW][MAX_HW];
int main() {
  scanf("%d %d", &H, &W);
  for (int i = 1; i <= H; i++) {
    for (int j = 1; j <= W; j++) {
      scanf("%d", &board[i][j]);
    }
  }
  for (int j = 1; j <= W; j++) {
    for (int i = 1; i <= H; i++) {
      if (board[i][j] == 1) {
        dp[i][j] = 0;
      } else {
        dp[i][j] = dp[i - 1][j] + 1;
      }
    }
  }
  int res = 0;
  for (int i = 1; i <= H; i++) {
    stack<pair<int, int> > st;
    for (int j = 1; j <= W; j++) {
      if (st.empty()) {
        st.push(make_pair(dp[i][j], j));
      } else if (st.top().first < dp[i][j]) {
        st.push(make_pair(dp[i][j], j));
      } else {
        int target = i;
        while (!st.empty() && st.top().first >= dp[i][j]) {
          pair<int, int> pre = st.top();
          st.pop();
          int area = pre.first * (j - pre.second);
          res = max(res, area);
          target = pre.second;
        }
        st.push(make_pair(dp[i][j], target));
      }
    }
  }
  cout << res << endl;
  return 0;
}
