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 ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
#define INF 1<<30
#define LINF 1LL<<60

struct Rectangle {
	ll h;
	ll p;  // position from left
};

ll max_area_of_histgram(ll n, vector<ll>& height) {
	stack<Rectangle> S;
	ll maxv = 0;

	for (int i = 0; i <= n; i++) {
		Rectangle rect;
		rect.h = height[i];
		rect.p = i;

		if (S.empty()) {
			S.push(rect);
		}
		else if (S.top().h < rect.h) {
			S.push(rect);
		}
		else if (S.top().h > rect.h) {
			ll now = i;
			while ((!S.empty()) && (S.top().h >= rect.h)) {
				Rectangle pre = S.top(); S.pop();
				ll area = 1LL * pre.h * (i - pre.p);
				maxv = max(maxv, area);
				now = pre.p;
			}
			rect.p = now;
			S.push(rect);
		}
	}
	return maxv;
}

int main() {
	cin.tie(0); ios::sync_with_stdio(false);
	ll N; cin >> N;
	vector<ll> dp(N + 1, 0);
	for (int i = 0; i < N;i++) cin >> dp[i];
	cout << max_area_of_histgram(N, dp) << endl;
	return 0;
}