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;
typedef long long int ll;
typedef vector<ll> vl;
typedef pair<ll, ll> PP;
#define rep(i, n) for (ll i = 0; i < ll(n); i++)
#define all(v) v.begin(), v.end()
bool chmin(ll& a, ll b) {
    if (b < a) {
        a = b;
        return 1;
    }
    return 0;
}
bool chmax(ll& a, ll b) {
    if (b > a) {
        a = b;
        return 1;
    }
    return 0;
}
const ll INF = 999999999999999;
const ll MOD = 1000000007;
const ll MAX_N = 500010;
ll a, b, c, d, e, f, p, t, x, y, z, q, m, n, r, h, k, w, l, ans, s;
ll histgram(vl B) {
    ll n = B.size(), ans = 0;
    vl A;
    A.push_back(0);
    rep(i, n) A.push_back(B[i]);
    A.push_back(0);
    n += 2;
    vl X(n, 0), Y(n, 0);

    deque<PP> DQ;

    rep(i, n) {
        while (!DQ.empty() && DQ.back().second > A[i]) {
            X[DQ.back().first] += i - DQ.back().first;
            DQ.pop_back();
        }
        DQ.push_back({i, A[i]});
    }

    reverse(all(A));

    DQ.clear();

    rep(i, n) {
        while (!DQ.empty() && DQ.back().second > A[i]) {
            Y[DQ.back().first] += i - DQ.back().first;
            DQ.pop_back();
        }
        DQ.push_back({i, A[i]});
    }

    reverse(all(Y));
    reverse(all(A));

    rep(i, n) {
        ans = max(ans, (X[i]+Y[i]-1) * A[i]);
    }
    
    return ans;
}
int main() {
    cin >> h >> w;
    vector<vl> dp(h, vl(w, 1));
    rep(i, h) {
        rep(j, w) {
            cin >> x;
            if (i > 0) dp[i][j] = dp[i - 1][j] + 1;
            if (x) dp[i][j] = 0;
        }
    }

    rep(i, h){
    	chmax(t, histgram(dp[i]));
    }

    cout << t << endl;
}

