Vika and Squares

Description

Vika has n jars with paints of distinct colors. All the jars are numbered from1 to
n and thei-th jar contains
ai liters of paint of colori.

Vika also has an infinitely long rectangular piece of paper of width
1, consisting of squares of size 1 × 1. Squares are numbered1,
2, 3 and so on. Vika decided that she will start painting squares one by one from left to right, starting from the square number1 and some arbitrary color. If the square
was painted in colorx, then the next square will be painted in colorx + 1. In case of
x = n, next square is painted in color1. If there is no more paint of the color Vika wants to use now, then she stops.

Square is always painted in only one color, and it takes exactly
1 liter of paint. Your task is to calculate the maximum number of squares that might be painted, if Vika chooses right color to paint the first square.

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 200 000) — the number of jars with colors Vika has.

The second line of the input contains a sequence of integers
a1, a2, ..., an (1 ≤ ai ≤ 109),
where ai is equal to the number of liters of paint in thei-th jar, i.e. the number of liters of colori
that Vika has.

Output

The only line of the output should contain a single integer — the maximum number of squares that Vika can paint if she follows the rules described above.

Sample Input

Input

5
2 4 2 3 3

Output

12

Input

3
5 5 5

Output

15

Input

6
10 10 10 1 10 10

Output

11

Hint

In the first sample the best strategy is to start painting using color
4. Then the squares will be painted in the following colors (from left to right): 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5.

In the second sample Vika can start to paint using any color.

In the third sample Vika should start painting using color number
5.

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define mymax 1111111111#define MAXN 200010
using namespace std;

long long int a[MAXN],ans,mi;
int main(){
int t,i,n,bz,len,m;
while(scanf("%d",&n)!=EOF){
for(i=1;i<=n;i++){
scanf("%I64d",&a[i]);
}
mi=mymax;
int k;
for(i=n;i>=1;i--){
if(a[i]<mi){
mi=a[i];
}
}
for(i=1;i<=n;i++){
a[i]-=mi;
}
len=0;
m=0;
for(i=1;i<=n;i++){
if(a[i])
len++;
else{
m=max(m,len);
len=0;
}
}
if(a
){
i=1;
while(a[i]){
len++;
i++;
}
m=max(m,len);
}
ans=mi*n+m;
printf("%I64d\n",ans);
}
return 0;
}