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
#pragma GCC optimize("Ofast")
#include <bits/stdc++.h>
using namespace std;
#define rep(i, n) for(int i = 0; i < (int)(n); ++i)
#define all(x) x.begin(),x.end()
#define ln '\n'
const double PI = acos(-1.0);
const long long MOD = 1000000007LL;
//const long long MOD = 998244353LL;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<long long, long long> pll;
template<class T> inline bool chmax(T &a, T b) { if (a < b) { a = b; return true;} return false; }
template<class T> inline bool chmin(T &a, T b) { if (a > b) { a = b; return true;} return false; }
///////////////////////////////////////////////////////////////////////////////////////////////////

int main() {
    ios::sync_with_stdio(false); cin.tie(nullptr);
    int N; cin >> N;
    vector<int> A(N);
    rep(i,N) cin >> A[i];

    ll ans = 0;
    stack<pll> sta;
    rep(i,N) {
        int a = i;
        while (!sta.empty() && sta.top().first > A[i]) {
            chmax(ans,sta.top().first*(i-sta.top().second));
            a = sta.top().second;
            sta.pop();
        }
        if (sta.empty() || sta.top().first < A[i]) sta.emplace(A[i],a); 
    }
    while (!sta.empty()) {
        chmax(ans, sta.top().first*(N-sta.top().second));
        sta.pop();
    }

    cout << ans << ln;
}
