A histogram is made of a number of contiguous bars, which have same width.

For a given histogram with $N$ bars which have a width of 1 and a height of $h_i$ = $h_1, h_2, ... , h_N$ respectively, find the area of the largest rectangular area.

Constraints

* $1 \leq N \leq 10^5$
* $0 \leq h_i \leq 10^9$

Input

The input is given in the following format.

$N$
$h_1$ $h_2$ ... $h_N$

Output

Print the area of the largest rectangle.

Examples

Input

8
2 1 3 5 3 4 2 1


Output

12


Input

3
2 0 1


Output

2

#include <bits/stdc++.h>
using namespace std;
#define rep(i,n,m) for(int (i)=(n);(i)<(m);(i)++)
#define rrep(i,n,m) for(int (i)=(n);(i)>(m);(i)--)
using ll = long long;
const ll mod = 1000000007;
const ll half = 500000004;

ll LRIH(vector<int> lis){

    vector<vector<int>> stk(0);

    ll ans = 0;
    ll one = 1;
    int N = lis.size();

    rep(i,0,N){

        if (stk.size() == 0){
            stk.push_back({lis[i],i});
        }else if (stk[stk.size()-1][0] < lis[i]){
            stk.push_back({lis[i],i});
        }else if (stk[stk.size()-1][0] > lis[i]){

            int lastpos = -1;

            while ( (stk.size()>0) && (stk[stk.size()-1][0] > lis[i])){

                int nh = stk[stk.size()-1][0];
                int np = stk[stk.size()-1][1];
                lastpos = np;
                stk.pop_back();

                ans = max(ans , (nh*one)*(i-np));
            }

            stk.push_back({lis[i] , lastpos});
        }
    }

    return ans;
}

int main(){

    int N;
    cin >> N;
    vector<int> h(N+1,0);
    rep(i,0,N){
        cin >> h[i];
    }
    cout << LRIH(h) << endl;
}
