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 <iostream>
#include <stack>

using namespace std;

struct rect {
    int pos, height;
};

int main() {

    int H, W;
    cin >> H >> W;

    int T[H][W];

    for (int i = 0; i < H; i++) {
        for (int j = 0; j < W; j++) {
            int c;
            cin >> c;
            if (i == 0) {
                T[i][j] = (c ? 0 : 1);
            } else {
                T[i][j] = (c ? 0 : T[i - 1][j] + 1);
            }
        }
    }

    //solve
    int ans = 0;

    for (int i = 0; i < H; i++) {
        stack<rect> stk;
        for (int j = 0; j <= W; j++) {
            int pos, height;
            pos = j;
            if (j == W) {
                height = 0;
            } else {
                height = T[i][j];
            }
            if (stk.empty() || stk.top().height < height) {
                stk.push((rect) {pos, height});
            } else if (stk.top().height > height) {
                while (!stk.empty() && stk.top().height >= height) {
                    rect &pre = stk.top();
                    stk.pop();
                    ans = max(ans, pre.height * (j - pre.pos));
                    pos = pre.pos;
                }
                stk.push((rect) {pos, height});
            }
        }
    }

    cout << ans << endl;
}
