#include<stdio.h>
#include<iostream>
#include<string.h>
#include<math.h>
#include<algorithm>
using namespace std;
const int N = 100005;
int min_dp[N][30];
long long a[N];
int L[N],R[N];
int MIN(int i,int j)
{
    if(a[i]>a[j]) return j;
    return i;
}
void init_RMQ(int n)
{
    for(int i=1; i<=n; i++)
    {
        min_dp[i][0]=i;
    }
    for(int j=1; (1<<j)<=n; j++)
    {
        for(int i=1; i+(1<<j)-1<=n; i++)
        {
            min_dp[i][j] = MIN(min_dp[i][j-1],min_dp[i+(1<<(j-1))][j-1]);
        }
    }
}

int MIN_RMQ(int l,int r)
{
    int k=0;
    while((1<<(k+1))<=(r-l+1)) k++;
    return MIN(min_dp[l][k],min_dp[r-(1<<k)+1][k]);
}
int binary(int value,int l,int r) ///找到最右边
{
    while(l<=r)
    {
        if(l==r) return l;
        int mid = (l+r)>>1;
        if(value<=a[MIN_RMQ(l,mid)])
        {
            l = mid+1;
        }
        else r = mid;
    }

}
int binary2(int value,int l,int r) ///找到最左边
{
    while(l<r)
    {
        if(l==r-1){   ///防止死循环,这里弄了好久
            if(a[r]<value) return r;  ///如果在r 并且a[r]<value 那么肯定r就是左边界
　　　　　　　return l;
        }
        int mid = (l+r)>>1;
        if(value<=a[MIN_RMQ(mid,r)])
        {
            r = mid-1;
        }
        else l = mid;
    }
    return l;
}
int main()
{
    int n;
    while(scanf("%d",&n)!=EOF&&n)
    {
        for(int i=1; i<=n; i++)
        {
            scanf("%lld",&a[i]);
        }
        init_RMQ(n);
        L[1]=1;
        R[n]=n;
        for(int i=1; i<=n; i++)
        {
            if(i!=n)
            {
                R[i] = binary(a[i],i+1,n);
                if(R[i]==n&&a[i]<=a[n]);
                else R[i]--;
            }
            if(i!=1)
            {
                L[i] = binary2(a[i],1,i-1);
                if(L[i]==1&&a[i]<=a[1]);
                else L[i]++;
            }
        }
        long long mx = -1;
        for(int i=1; i<=n; i++)
        {
            if((R[i]-L[i]+1)*a[i]>mx) mx = (R[i]-L[i]+1)*a[i];
        }
        printf("%lld\n",mx);
    }

}

#include <stdio.h>
#include <iostream>
#include <string.h>
#include <math.h>
#include <algorithm>
using namespace std;
const int N = 100005;
long long a[N],L[N],R[N]; ///L[i]记录i点能够到达最左边的位置，R[i]同理
int main()
{
    int n;
    while(scanf("%d",&n)!=EOF&&n){
        for(int i=1;i<=n;i++){
            scanf("%lld",&a[i]);
        }
        L[1]=1;
        R[n]=n;
        for(int i=2;i<=n;i++){ ///先求出每个坐标最左边能到达的位置
            int t = i;
            while(t>1&&a[i]<=a[t-1]){
                t = L[t-1];       ///这步相当巧妙,直接实现了跳转，如果a[t-1]不小于a[i]
                                  ///的话，我们可以断定a[t-1]的最左边肯定包含a[i]的最左边.直接跳过中间的点
                                  ///时间复杂度就肯定没有O(n*n)了,应该是O(n*k) k是个比较小的数
            }
            L[i] = t;
        }
        for(int i=n-1;i>=1;i--){ ///找最右边
            int t =i;
            while(t<n&&a[i]<=a[t+1]){
                t = R[t+1];
            }
            R[i] = t;
        }
        long long mx = -1;
        for(int i=1;i<=n;i++){
            if((R[i]-L[i]+1)*a[i]>mx) mx = (R[i]-L[i]+1)*a[i];
        }
        printf("%lld\n",mx);
    }
    return 0;
}