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 int long long int
#define MOD(x) ((x % MOD_N) + MOD_N) % MOD_N
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define FORE(i,a,b) for(int i=(a);i<=(b);++i)
#define RFOR(i,a,b) for(int i=(b)-1;i>=(a);--i)
#define RFORE(i,a,b) for(int i=(b);i>=(a);--i)
#define REP(i,n) FOR(i,0,n)
#define RREP(i,n) RFOR(i,0,n)
#define ALL(c) (c).begin(),(c).end()
#define RALL(c) (c).rbegin(),(c).rend()
#define SORT(c) sort(ALL(c))
#define RSORT(c) sort(RALL(c))
#define SZ(c) (int)((c).size())
#define EACH(i,v) for(auto i=v.begin();i!=v.end();++i)
#define REACH(i,v) for(auto i=v.rbegin();i!=v.rend();++i)
#define LB(c,x) distance((c).begin(),lower_bound(ALL(c),x))
#define UB(c,x) distance((c).begin(),upper_bound(ALL(c),x))
#define COUNT(c,x) (lower_bound(ALL(c),x)-upper_bound(ALL(c),x))
#define UNIQUE(c) SORT(c); (c).erase(unique(ALL(c)),(c).end());
#define COPY(c1,c2) copy(ALL(c1),(c2).begin())
#define EXIST(s,e) (bool)((s).find(e)!=(s).end())
#define PB push_back
#define MP make_pair
#define DUMP(x)  cerr<<#x<<" = "<<(x)<<endl;
#define NL cerr<<endl;
using namespace std;
template<typename T,typename U> using P=pair<T,U>;
template<typename T> using V=vector<T>;
template<typename T>bool chmax(T&a,T b){if(a<b){a=b;return true;}return false;}
template<typename T>bool chmin(T&a,T b){if(a>b){a=b;return true;}return false;}
template<typename T>T sum(vector<T>&v){return accumulate(ALL(v),T());}
template<typename T>T sum(vector<T>&v,int a,int b){return accumulate(v.begin()+a,v.begin()+b,T());}
template<typename T>T max(vector<T>&v){return *max_element(ALL(v));}
template<typename T>T min(vector<T>&v){return *min_element(ALL(v));}
template<typename T>T max_index(vector<T>&v){return distance((v).begin(),max_element(ALL(v)));}
template<typename T>T min_index(vector<T>&v){return distance((v).begin(),min_element(ALL(v)));}

struct edge { int to, cost; };

template<typename T>auto&operator<<(ostream&s,const vector<T>&v){s<<"[";bool a=1;for(auto e:v){s<<(a?"":" ")<<e;a=0;}s<<"]";return s;}
template<typename T,typename U>auto&operator<<(ostream&s,const pair<T,U>&p){s<<"("<<p.first<<","<<p.second<<")";return s;}
template<typename T>auto&operator<<(ostream&s,const set<T>&st){s<<"{";bool a=1;for(auto e:st){s<<(a?"":" ")<<e;a=0;}s<<"}";return s;}
template<typename T,typename U>auto&operator<<(ostream&s,const map<T,U>&m){s<<"{";bool a=1;for(auto e:m){s<<(a?"":" ")<<e.first<<":"<<e.second;a=0;}s<<"}";return s;}

const int INF = 1e18;
const int MOD_N = 1e9+7;

signed main()
{
   int N; cin >> N;
   V<int> h(N+2);
   REP(i, N) {
      cin >> h[i+1];
   }
   deque<P<int,int>> deq;
   int ans = 0;
   REP(i, N+2) {
      while (!deq.empty() && deq.back().first >= h[i]) {
         int m = deq.back().first;
         deq.pop_back();
         int id = deq.back().second;
         chmax(ans , m * (i - id - 1));
      }
      deq.push_back(MP(h[i], i));
   }
   cout << ans << endl;

   return 0;
}

