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)

// In the Name of God

#include <bits/stdc++.h>

using namespace std;

#define	ll long long
#define ull unsigned long long
#define ld long double
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define sz(a) int(a.size())
#define all(v) v.begin(), v.end()
#define bpc(v) __builtin_popcountll(v)
#define itr iterator
#define rep(i, a, b) for (int i = a; i <= b; ++i)
#define per(i, a, b) for (int i = a; i >= b; --i)
#define ub upper_bound
#define lb lower_bound

const int N = 1e5 + 5;
const int inf = 1e9 + 1;
const int mod = 1e9 + 7;
const double eps = 1e-15;
const int pw = 257;

int n, a[N];
ll sum;
multiset<int> s;

int main() {
	#ifdef Madi
	freopen(".in", "r", stdin);
	freopen(".out", "w", stdout);
	#endif

	ios_base :: sync_with_stdio(false); cin.tie(NULL);

	cin >> n;
	rep(i, 1, n)
		cin >> a[i];

	rep(i, 1, n) {
		s.insert(a[i]);
		sum += a[i];
	}

	if (sum % (n * 1ll * (n + 1) / 2) != 0)
		return cout << "NO", 0;

	int c = sum / (n * 1ll * (n + 1) / 2);
	per(i, n, 1) {
		rep(i, 1, c) {
			int mx = *s.rbegin();
			if (*s.rbegin() < i)
				return cout << "NO", 0;
			s.erase(s.find(mx));
			mx -= i;
			s.insert(mx);
		}
	}

	cout << "YES";
	return 0;
}