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<iomanip>
#include<limits>
#include<thread>
#include<utility>
#include<iostream>
#include<string>
#include<algorithm>
#include<set>
#include<map>
#include<vector>
#include<stack>
#include<queue>
#include<cmath>
#include<numeric>
#include<cassert>
#include<random>
#include<chrono>
#include<unordered_map>
#include<fstream>
#include<list>
#include<functional>
using namespace std;
typedef unsigned long long int ull;
typedef long long int ll;
typedef pair<ll,ll> pll;
typedef pair<int,int> pi;
typedef pair<double,double> pd;
typedef pair<double,ll> pdl;
#define F first
#define S second
const ll E=1e18+7;
const ll MOD=1000000007;

ll Largest_Rectangle_inHistogram(const vector<ll> &A){
    ll mx=0;
    stack<pll> s;
    for(ll i=0;i<A.size();i++){
        ll mi=i;
        while(!s.empty() && s.top().F>A[i]){
            pll w=s.top(); s.pop();
            mx=max(mx,w.F*(i-w.S));
            mi=w.S;
        }
        s.push({A[i],mi});
    }
    while(!s.empty()){
        pll w=s.top(); s.pop();
        mx=max(mx,w.F*((ll)A.size()-w.S));
    }
    return mx;
}


ll Largest_Rectangle(const vector<vector<bool>> &A){
    vector<vector<ll>> a(A.size(),vector<ll>(A[0].size(),0));
    ll count;
    for(int i=0;i<A[0].size();i++){
        count=0;
        for(ll h=0;h<A.size();h++){
            if(A[h][i]){count++;}
            else{count=0;}
            a[h][i]=count;
        }
    }
    ll mx=0;
    stack<pll> s;
    for(int i=0;i<A.size();i++){
        for(int t=0;t<A[0].size();t++){
            ll mi=t;
            while(!s.empty() && s.top().F>a[i][t]){
                pll w=s.top(); s.pop();
                mx=max(mx,w.F*(t-w.S));
                mi=w.S;
            }
            s.push({a[i][t],mi});
        }
        while(!s.empty()){
            pll w=s.top(); s.pop();
            mx=max(mx,w.F*((ll)A[0].size()-w.S));
        }
    }
    return mx;
}

ll Largest_Square(const vector<vector<bool>> &A){
    vector<vector<ll>> a(A.size(),vector<ll>(A[0].size(),0));
    ll count;
    for(int i=0;i<A[0].size();i++){
        count=0;
        for(ll h=0;h<A.size();h++){
            if(A[h][i]){count++;}
            else{count=0;}
            a[h][i]=count;
        }
    }
    ll mx=0;
    stack<pll> s;
    for(int i=0;i<A.size();i++){
        for(int t=0;t<A[0].size();t++){
            ll mi=t;
            while(!s.empty() && s.top().F>a[i][t]){
                pll w=s.top(); s.pop();
                ll x=min(w.F,t-w.S);
                mx=max(mx,x*x);
                mi=w.S;
            }
            s.push({a[i][t],mi});
        }
        while(!s.empty()){
            pll w=s.top(); s.pop();
            ll x=min(w.F,(ll)A[0].size()-w.S);
            mx=max(mx,x*x);
        }
    }
    return mx;
}


int main(){
    ll n;
    cin>>n;
    vector<ll> a(n);
    for(int i=0;i<n;i++){cin>>a[i];}
    cout<<Largest_Rectangle_inHistogram(a)<<endl;
    
    return 0;
}

