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
//Largest Square
#include <iostream>
#include <stack>
#include <algorithm>

using namespace std;

static const int MAX = 1400;

struct Rectangle {
public:
    int height, pos;
    Rectangle(int height, int pos): height(height), pos(pos) {}
};

int G[MAX][MAX];
int H, W;

int getLargestRectangle() {
    stack<Rectangle> S;
    int dp[H][W];
    int maxArea = 0;

    for(int i = 0; i < H; i++) {
        for(int j = 0; j < W; j++) {
            dp[i][j] = (G[i][j] + 1) % 2;
            if(i != 0 && dp[i][j] != 0) dp[i][j] += dp[i - 1][j];
        }
    }

    for(int i = 0; i < H; i++) {
        for(int j = 0; j < W; j++) {
            Rectangle rect = Rectangle(dp[i][j], j);
            if(S.empty()) S.push(rect);
            else {
                if(S.top().height < rect.height) {
                    S.push(rect);
                } else if(S.top().height > rect.height) {
                    int target = j;
                    while(!S.empty() && S.top().height > rect.height) {
                        maxArea = max(maxArea, (j - S.top().pos) * S.top().height);
                        target = S.top().pos;
                        S.pop();
                    }
                    S.push(Rectangle(rect.height, target));

                }
            }
        }
        while(!S.empty()) {
            maxArea = max(maxArea, (W - S.top().pos) * S.top().height);
            S.pop();
        }
    }
    return maxArea;

}

int main() {
    int c;
    cin >> H >> W;
    for(int i = 0; i < H; i++) {
        for(int j = 0; j < W; j++) cin >> G[i][j];
    }

    cout << getLargestRectangle() << endl;
}

