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 ll;
#define REP(i,n) FOR(i,0,n)
#define FOR(i,a,b) for(ll i=a;i<b;i++)
#define PB push_back
#define LB lower_bound
#define UB upper_bound
#define PQ priority_queue
#define UM unordered_map
#define ALL(a) (a).begin(),(a).end()
typedef vector<ll> vi;
typedef vector<vector<ll>> vvi;
const ll INF = (1ll << 60);
typedef pair<ll,ll> pii;

ll largestAreaHistogram(vi v){
  ll N=v.size();
  stack<pii> st; st.push({0,v[0]});
  ll ans=v[0];
  FOR(i,1,N) {
    ll tmp=i;
    while(st.size()&&st.top().second>v[i]) {
      ans=max(ans,st.top().second*(i-st.top().first));
      tmp=st.top().first;
      st.pop();
    }
    st.push({tmp,v[i]});
  }
  while(st.size()) {
    ans=max(ans,st.top().second*(N-st.top().first));
    st.pop();
  }
  return ans;
}

int main(){
  ll H,W; cin>>H>>W;
  vvi c(H,vi(W));
  REP(h,H) REP(w,W) cin>>c[h][w];
  vvi dp(H,vi(W));
  REP(w,W) if(c[0][w]==0) dp[0][w]=1; else dp[0][w]=0;
  FOR(h,1,H) REP(w,W) if(c[h][w]==0) dp[h][w]=dp[h-1][w]+1; else dp[h][w]=0;
  ll ans=0; REP(h,H) ans=max(ans,largestAreaHistogram(dp[h]));
  cout<<ans<<endl;
}