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>
using namespace std;
int  n,a[100005],b[100005],sum=0,ans=0;
inline int read(){
	char c=getchar();
	int x=0;
	for(;c>'9'||c<'0';) c=getchar();
	for(;c>='0'&&c<='9';){
		x=(x<<3)+(x<<1)+c-'0';
		c=getchar();
	}
	return x;
}
int main(){
	n=read();
	for(int i=1;i<=n;i++) a[i]=read(),sum+=a[i];
	if(sum%(n*(n+1)/2)){
		puts("NO");
		return 0;
	}
	sum/=(n*(n+1))/2;
	b[1]=a[1]-a[n];
	for(int i=2;i<=n;i++) b[i]=a[i]-a[i-1];
	for(int i=1;i<=n;i++)
		if((sum-b[i]<0)||(sum-b[i])%n){
			puts("NO"); 
			return 0;
		}
		else ans+=(sum-b[i])/n;
	if(ans<sum) puts("NO");
	else puts("YES");
	return 0;
}