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>
#define FOR(i,a,b) for(int i= (a); i<((int)b); ++i)
#define RFOR(i,a) for(int i=(a); i >= 0; --i)
#define FOE(i,a) for(auto i : a)
#define ALL(c) (c).begin(), (c).end()
#define RALL(c) (c).rbegin(), (c).rend()
#define DUMP(x)  cerr << #x << " = " << (x) << endl;
#define SUM(x) std::accumulate(ALL(x), 0LL)
#define MIN(v) *std::min_element(v.begin(), v.end())
#define MAX(v) *std::max_element(v.begin(), v.end())
#define EXIST(v,x) (std::find(v.begin(), v.end(), x) != v.end())
#define BIT(n) (1LL<<(n))
#define UNIQUE(v) v.erase(unique(v.begin(), v.end()), v.end());

typedef long long LL;
template<typename T> using V = std::vector<T>;
template<typename T> using VV = std::vector<std::vector<T>>;
template<typename T> using VVV = std::vector<std::vector<std::vector<T>>>;
template<class T> inline T ceil(T a, T b) { return (a + b - 1) / b; }
template<class T> inline void print(T x) { std::cout << x << std::endl; }
template<class T> inline bool inside(T y, T x, T H, T W) {return 0 <= y and y < H and 0 <= x and x < W; }
inline double distance(double y1, double x1, double y2, double x2) { return sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2)); }

const int INF = 1L << 30;
const double EPS = 1e-9;
const std::string YES = "YES", Yes = "Yes", NO = "NO", No = "No";
const std::vector<int> dy = { 0, 1, 0, -1 }, dx = { 1, 0, -1, 0 };    // 4近傍（右, 下, 左, 上）

using namespace std;



int main() {
    int N;
    cin >> N;

    vector<LL> height(N);
    FOR(i, 0, N) {
        cin >> height[i];
    }

    // L
    vector<int> L(N);
    {
        int t = 0; // スタックのサイズ
        int st[N];
        for (int i = 0; i < N; ++i) {
            while (t > 0 and height[st[t - 1]] >= height[i]) {
                t--;
            }
            L[i] = (t == 0 ? 0 : (st[t - 1] + 1));
            st[t++] = i;
        }
    }

    // R
    vector<int> R(N);
    {
        int t = 0;
        int st[N];
        for (int i = N - 1; i >= 0; i--) {
            while (t > 0 and height[st[t - 1]] >= height[i]) {
                t--;
            }
            R[i] = (t == 0 ? N : st[t - 1]);
            st[t++] = i;
        }
    }

    long long ans = 0;
    for (int i = 0; i < N; i++) {
        ans = max(ans, (long long)height[i] * (R[i] - L[i]));
    }

    print(ans);
    return 0;
}
