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 <iostream>
#include <string>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <map>
#include <set>
#include <algorithm>
#include <queue>
#include <stack>
#include <functional>
#include <bitset>
#include <assert.h>
using namespace std;
typedef long long ll;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef vector<char> vc;
typedef vector<string> vs;
typedef vector<bool> vb;
typedef vector<double> vd;
typedef pair<ll,ll> P;
typedef vector<P> vpl;
typedef tuple<ll,ll,ll> tapu;
#define rep(i,n) for(ll i=0; i<(n); i++)
#define REP(i,a,b) for(int i=(a); i<(b); i++)
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
const int inf = 1<<30;
const ll linf = 1LL<<62;
const int MAX = 510000;
ll dy[8] = {0,1,0,-1,1,-1,1,-1};
ll dx[8] = {1,0,-1,0,1,-1,-1,1};
const double pi = acos(-1);
const double eps = 1e-7;
template<typename T1,typename T2> inline bool chmin(T1 &a,T2 b){
	if(a>b){
		a = b; return true;
	}
	else return false;
}
template<typename T1,typename T2> inline bool chmax(T1 &a,T2 b){
	if(a<b){
		a = b; return true;
	}
	else return false;
}
template<typename T1,typename T2> inline void print2(T1 a, T2 b){cout << a << " " << b << "\n";}
template<typename T1,typename T2,typename T3> inline void print3(T1 a, T2 b, T3 c){
	cout << a << " " << b << " " << c << "\n";
}
const int mod = 1e9 + 7;
//const int mod = 998244353;


int main(){
    ll n; cin >> n;
    vl h(n); rep(i,n) cin >> h[i];
    ll ans = 0;
    stack<P> st;
    rep(i,n){
        if(st.empty()) st.emplace(h[i],i);
        else{
            ll mn = inf;
            while(!st.empty()){
                P p = st.top();
                if(h[i] > p.first) break;
                st.pop();
                chmax(ans, p.first * (i-p.second));
                chmin(mn,p.second);
            }
            st.emplace(h[i],min(i,mn));
        }
    }
    while(!st.empty()){
        P p = st.top();
        chmax(ans, p.first * (n-p.second));
        st.pop();
    }
    cout << ans << endl;
}
