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)

//code by lynmisakura.wish to be accepted!
/****************************/
#include<iostream>
#include<iomanip>
#include<math.h>
#include<vector>
#include<string>
#include<stack>
#include<queue>
#include<map>
#include<algorithm>
#include<bitset>
#include <climits>
#include<set>
#include<bitset>
using namespace std;
/***************************/
typedef long long ll;
typedef long long ijt;
typedef vector<int> vi;
typedef vector<long long> vl;
typedef pair<int, int> pi;
typedef vector<pair<int, int>> vpi;

const long long INF = 1LL << 55;

#define itn int
#define endl '\n'
#define pb push_back
#define mp make_pair
#define ss second
#define ff first
#define dup(x,y) ((x) + (y) - 1)/(y)
#define mins(x,y) x = min(x,y)
#define maxs(x,y) x = max(x,y)
#define all(x) (x).begin(),(x).end()
#define Rep(n) for(int i = 0;i < n;i++)
#define rep(i,n) for(int i = 0;i < n;i++)
#define rrep(i,n) for(int i = n - 1;i >= 0;i--)
#define UNIQUE(v) v.erase( unique(v.begin(), v.end()), v.end() )

ll gcd(ll a, ll b) { return b ? gcd(b, a%b) : a; }
ll qp(ll a, ll b) { ll ans = 1; do { if (b & 1)ans = 1ll * ans*a; a = 1ll * a*a; } while (b >>= 1); return ans; }
ll qp(ll a, ll b, int mo) { ll ans = 1; do { if (b & 1)ans = 1ll * ans*a%mo; a = 1ll * a*a%mo; } while (b >>= 1); return ans; }

#define _GLIBCXX_DEBUG
#define debug(x) cerr << #x << " = " << (x) << " (L" << __LINE__ << ")" << " " << __FILE__ << endl;

int main(void) {
	ijt n;cin >> n;
	vector<ijt> a(n+1);
	ijt sum = 0;
	rep(i, n) {
		cin >> a[i+1];
		sum += a[i+1];
	}
	ijt m;
	if (sum % (n*(n + 1) / 2) != 0) {
		cout << "NO" << endl;
		return 0;
	}
	else {
		m = sum / (n*(n+1)/2);
	}

	ijt xsum = 0;
	for (int i = 1; i <= n; i++) {
		if (i != n) {
			ijt d = a[i+1] - a[i];
			if (((n-1)* m + d) % n != 0) {
				cout << "NO" << endl;return 0;
			}
			else {
				xsum += ((n-1)*m + d)/n;
			}
		}
		else {
			ijt d = a[1] - a[n];
			if (((n - 1)* m + d) % n != 0) {
				cout << "NO" << endl; return 0;
			}
			else {
				xsum += ((n - 1)*m + d) / n;
			}
		}
	}
	if (xsum != m*(n-1)) {
		cout << "NO" << endl;
	}
	else {
		cout << "YES" << endl;
	}
	return 0;
}