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<climits>
#include<algorithm>
#include<stack>
using namespace std;
#define FOR(i,a,b) for(int i=(a);i<(b);i++)
#define REP(i,n) for(int i=0;i<(n);i++)
#define INF INT_MAX/3
#define PII pair<int,int>
#define MPII(a,b) make_pair((a),(b))

int M[1400][1400];
int dp[1400][1401] = {};

int main()
{
	int ans = 0;
	int H, W;
	cin >> H >> W;
	REP(i, H)REP(j, W)cin >> M[i][j];
	REP(j, W)if (M[0][j] == 0)dp[0][j] = 1;
	FOR(i, 1, H)REP(j, W)if (M[i][j] == 0)dp[i][j] = dp[i - 1][j] + 1;
	REP(i, H)
	{
		stack<PII> stk;
		REP(j, W + 1)
		{
			if (stk.empty() || stk.top().second < dp[i][j])
			{
				stk.push(MPII(j, dp[i][j]));
			}
			else if (stk.top().second > dp[i][j])
			{
				int temp;
				while (stk.top().second > dp[i][j])
				{
					int rect = (j - stk.top().first)*stk.top().second;
					ans = max(ans, rect);
					temp = stk.top().first;
					stk.pop();
					if (stk.empty())break;
				}
				stk.push(MPII(temp, dp[i][j]));
			}
		}
	}
	cout << ans << endl;
	return 0;
}