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;
template <typename T>
ostream& operator<<(ostream& os, vector<T>& vec) {
  os << "{";
  for (int i = 0; i < vec.size(); i++)
    os << vec[i] << (i + 1 == vec.size() ? "" : ", ");
  os << "}";
  return os;
}
void dump_func() { cout << endl; }
template <class Head, class... Tail>
void dump_func(Head&& head, Tail&&... tail) {
  cout << head;
  if (sizeof...(Tail) == 0) {
    cout << " ";
  } else {
    cout << ", ";
  }
  dump_func(std::move(tail)...);
}
template <class T>
inline void chmax(T& a, T const& b) {
  a = max(a, b);
}
template <class T>
inline void chmin(T& a, T const& b) {
  a = min(a, b);
}
using ll = long long;
using P = pair<int, int>;
using Pl = pair<ll, ll>;
using vi = vector<int>;
using vvi = vector<vi>;
using vl = vector<ll>;
using vvl = vector<vl>;
const int mod = 1e9 + 7, INF = 1 << 30;
const double EPS = 1e-12, PI = 3.1415926535897932384626;
const ll lmod = 1e9 + 7, LINF = 1LL << 60;
const int MAX_N = 200005;
using sz = ll;
long long maximum_rectangle(const std::vector<sz>& h) {
  int n = h.size();
  std::stack<sz> s;
  std::vector<sz> L(n), R(n);
  for (int i = 0; i < n; i++) {
    while (s.size() && h[s.top()] >= h[i]) s.pop();
    L[i] = s.size() ? (s.top() + 1) : 0;
    s.push(i);
  }
  while (s.size()) s.pop();
  for (int i = n - 1; i >= 0; i--) {
    while (s.size() && h[s.top()] >= h[i]) s.pop();
    R[i] = s.size() ? s.top() : n;
    s.push(i);
  }
  long long res = 0;
  for (int i = 0; i < n; i++) {
    res = std::max(res, (long long)h[i] * (R[i] - L[i]));
  }
  return res;
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int N;
  cin >> N;
  vector<sz> h(N);
  for (int i = 0; i < N; i++) cin >> h[i];
  cout << maximum_rectangle(h) << endl;
  return 0;
}
