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: Accepted)

#include <bits/stdc++.h>
#define dbug(x) cout<<#x<<"="<<x<<endl
using namespace std;
template <typename T> void read(T &t) {
	t=0; char ch=getchar(); int f=1;
	while ('0'>ch||ch>'9') { if (ch=='-') f=-1; ch=getchar(); }
	do {(t*=10)+=ch-'0';ch=getchar();} while ('0'<=ch&&ch<='9'); t*=f;
}
typedef long long ll;
const int maxn=(1e5)+10;
int n,a[maxn],mn=(1e9),mx;
ll s1,k,s2;
int main() {
	//freopen("1.txt","r",stdin);
	read(n);
	for (int i=1;i<=n;i++) {
		read(a[i]),s1+=a[i];
		mn=min(mn,a[i]);
		mx=max(mx,a[i]);
	}
	s2=(ll)n*(n+1)/2;
	if (s1%s2!=0) { printf("NO\n"); return 0; }
	k=s1/s2;
	if (mn<k||mx>k*n) { printf("NO\n"); return 0; }
	for (int i=1;i<=n;i++)
		a[i]=(a[i]-k*i%n+n)%n;
	for (int i=1;i<=n;i++)
		if (a[i]!=a[1]) { printf("NO\n"); return 0; }
	printf("YES\n");
	return 0;
}