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 <algorithm>
# include <vector>
# include <string>
# include <set>
# include <map>
# include <cmath>
# include <iomanip>
# include <functional>
# include <utility>
# include <stack>
# include <queue>
# include <list>
# include <bitset>
# include <complex>
#include<limits.h>
#include<unordered_map>
#include<unordered_set>
#include<deque>
#include<cstdio>
using namespace std;
typedef long long int ll;
const int N = 1000000;
const int mod = 1000000007;
const int INF = 1 << 30;
#define rep(i,n) for(ll i=(ll)0;i<(ll)n;++i)
#define ALL(x) x.begin(),x.end()
#define pp pair<ll,ll>
#define fi first
#define se second
#define pb push_back
ll ppow(ll x, ll n) {
	ll ans = 1;
	while (n > 0) {
		if ((n & 1) == 1)ans = ans * x;
		x = x * x;
		n >>= 1;
		x %= mod;
		ans %= mod;
	}
	return ans;
}
string YN(bool b) { return(b ? "YES" : "NO"); }
string yn(bool b) { return(b ? "Yes" : "No"); }
ll sz, fact[N], inv[N];
void setinv() {
	fact[0] = 1; inv[0] = 1;
	for (int i =1; i <= sz; i++) {
		fact[i] = (fact[i - 1] * i) % mod;  //階乗を求める
		inv[i] = ppow(fact[i], (ll)mod - 2) % mod; // フェルマーの小定理で逆元を求める
	}
}
ll conv(ll r) {//nCr
	return fact[sz] * inv[r] % mod * inv[sz - r] % mod;
}
ll n, v[100000],ct=0,sum=0,kaisuu,sum2;
int main() {
	cin >> n;
	rep(i, n) { cin >> v[i]; sum += v[i]; sum2 += i + 1; }
	kaisuu = sum / sum2;
	bool b=1;
	rep(i,n) {
		if (v[(i + n - 1) % n] >= v[i]) {
			rep(j, kaisuu + 1) { if (4 * j - (kaisuu - j) == v[(i + n - 1) % n] - v[i])ct += j; }
		}
		else b = b&&v[i] - v[(i + n - 1) % n] == kaisuu;
	}
	cout<<YN(b&&sum%sum2==0&&abs(ct-kaisuu)%5==0) << endl;
	return 0;
}