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 <bits/stdc++.h>
#define int long long
#define uint unsigned int
#define rep(i, a, n) for (int i = a; i < n; i++)
#define all(a) (a).begin(), (a).end()
#define sz(a) (a).size()
#define PB push_back
#define EB emplace_back
#define MP make_pair
#define MT make_tuple
#define dump(x) cerr << #x << " = " << (x) << endl;
#define dumpi(i, x) cerr << string((i), ' ') << #x << " = " << (x) << endl;
#define debug(x) cerr << #x << " = " << (x) << " (L" << __LINE__ << ")" << " " << __FILE__ << endl;
using namespace std;
using pii = pair<int, int>;
constexpr int MOD = 1000000007;
constexpr int INF = 1LL << 30;
constexpr double EPS = 1e-10;

void no() {
  cout << "NO" << endl;
  exit(0);
}

int N;
int A[100010];
int d[100010];

signed main() {
  cin.tie(0);
  ios_base::sync_with_stdio(false);
  cout << fixed << setprecision(10);
  
  cin >> N;
  int sum = 0;
  rep(i, 0, N) {
    cin >> A[i];
    sum += A[i];
  }
  int tmp = N*(N+1)/2;
  if (sum % tmp != 0) no();
  int k = sum / tmp;
  d[0] = A[0]-A[N-1];
  rep(i, 1, N) {
    d[i] = A[i]-A[i-1];
  }
  int sum2 = 0;
  rep(i, 0, N) {
    int x = d[i]+k*(N-1);
    if (x < 0 || x % N != 0) no();
    x /= N;
    sum2 += k-x;
  }
  cout << (sum2 == k ? "YES" : "NO") << endl;

  return 0;
}