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;

typedef long long int ll;
typedef pair<int,int> P;

#define rep(i,n) for(int i=0;i<(n);++i)
#define repi(i,a,b) for(int i=int(a);i<(b);i++)
#define repr(i,b,a)for(int i=b;i>=a;i--)
#define all(x) x.begin(),x.end()

const ll mod = 1e9+7;
const ll INF = 1e9;

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
int dx[4]={1,0,-1,0};
int dy[4]={0,1,0,-1};

int main()
{
	int n;
	cin>>n;
	vector<ll> h(n);
	rep(i,n)cin>>h[i];

	stack<ll> st;
	vector<ll> L(n),R(n);

	L[0] = 0;
	rep(i,n){
		while(st.size() && h[st.top()]>=h[i])st.pop();
		L[i] = (st.empty()?0:st.top()+1);
		st.push(i);
	}
	while(st.size())st.pop();
	R[n-1]=0;
	repr(i,n-1,0){
		while(st.size() && h[st.top()]>=h[i])st.pop();
		R[i] = (st.empty()?n:st.top());
		st.push(i);
	}
	while(st.size())st.pop();

	ll ans=0;
	rep(i,n){
		ans = max(ans,h[i]*(R[i]-L[i]));
	}
	cout<<ans<<endl;

	return 0;
}
