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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

/*
 * Author : Andrew J
*/
#include <bits/stdc++.h>
using namespace std;
#define rall(x) x.rbegin() , x.rend()
#define all(x) x.begin() , x.end()
#define fi first
#define se second
#define pb push_back
using ll = long long;
using ii = pair <int , int>;
void in(int &x);
// const int N = ;
#define singleCase
// #define multipleCase
int n;

int modular(int x , int m) {
   return ((x % m) + m) % m;
}

void work() {
   in(n);
   vector <int> v(n);
   for (int i = 0 ; i < n ; i++) {
      in(v[i]);
   }
   set <int> s;
   for (int i = 0 ; i < n - 1 ; i++) {
      s.insert(modular(v[i] - v[i + 1] , n));
   }
   s.insert(modular(v[n - 1] - v[0] , n));
   puts(s.size() == 1 ? "YES" : "NO");
   return;
}

int main() {
   #ifdef multipleCase
      int nQ; in(nQ);
      while (nQ--)
         work();
   #endif
   #ifdef singleCase
      work();
   #endif
   return 0;
}

void in(int &x) {
	bool neg = false;
	register int c;
	x = 0;
	c = getchar();
	if(c == '-') {
		neg = true;
		c = getchar();
	}
	for (; c > 47 && c < 58; c = getchar())
	  x = (x << 1) + (x << 3) + c - 48;
	if (neg) x *= -1;
}