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: Runtime Error)

#include <iostream>
using namespace std;
long long diff[100002];
long long k[100002];
int main(){
int N;
cin >>N;

int a=0;
for(int i=0;i<N;i++){
cin>>k[i];
diff[i]+=k[i];
diff[(i+1)%N]-=k[i];
}
long long min=diff[0];
for(int i=0;i<N;i++){
if(diff[i]%(N-1)!=0){
cout<<"NO"<<endl;
return 0;
}
if (min>diff[i])min=diff[i];
}
if (min<0){
for(int i=0;i<N;i++){
diff[i]-=min;
}}
for(int i=0;i<N;i++){
k[0]-=(N-i)*diff[i]/(N-1);
}
if(k[0]>=0&&k[0]%(N*(N+1)/2)==0){
cout<<"YES"<<endl;

return 0;
}
cout<<"NO"<<endl;
return 0;
}