Problem:
Score : 500 points

Problem Statement

There are N boxes arranged in a circle. The i-th box contains A_i stones.

Determine whether it is possible to remove all the stones from the boxes by repeatedly performing the following operation:

Select one box. Let the box be the i-th box. Then, for each j from 1 through N, remove exactly j stones from the (i+j)-th box. Here, the (N+k)-th box is identified with the k-th box.

Note that the operation cannot be performed if there is a box that does not contain enough number of stones to be removed.

Constraints

1 ≦ N ≦ 10^5

1 ≦ A_i ≦ 10^9

Input

The input is given from Standard Input in the following format:

N
A_1 A_2 … A_N

Output

If it is possible to remove all the stones from the boxes, print YES. Otherwise, print NO.

Sample Input 1

5
4 5 1 2 3

Sample Output 1

YES

All the stones can be removed in one operation by selecting the second box.

Sample Input 2

5
6 9 12 10 8

Sample Output 2

YES

Sample Input 3

4
1 2 3 1

Sample Output 3

NO

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <cstdio>
#include <string.h>
using namespace std;


int main(){
    int n;
    while(cin>>n){
        long long evens=0, odds=0;
        while(n--){
            long long a;
            scanf("%lld", &a)==1;
            if(a%2==0) evens++;
            else odds++;
        }
        if(odds%2==0){
            odds>>=1;
            evens+=odds;
            evens%=2;
            if(evens==0 || (odds==0&&evens==1))  printf("YES\n");
            else printf("NO\n");
        }
        else{
            /*odds>>=1;
            evens+=odds;
            odds=1;
            if(evens%2==0)  printf("YES\n");*/
            printf("NO\n");
        }
    }
return 0;
}