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: Accepted)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<ll, ll> P;
#define DUMP(x) cout << #x << " = " << (x) << endl;
#define FOR(i, m, n) for (ll i = m; i < n; i++)
#define IFOR(i, m, n) for (ll i = n - 1; i >= m; i--)
#define REP(i, n) FOR(i, 0, n)
#define IREP(i, n) IFOR(i, 0, n)
#define FOREACH(x, a) for (auto&(x) : (a))
#define ALL(v) (v).begin(), (v).end()
#define SZ(x) ll(x.size())

void yes() {
  cout << "YES" << endl;
  exit(0);
}

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

int main() {
  ll n;
  cin >> n;
  vector<ll> a(n);
  REP(i, n) cin >> a[i];
  ll sum = 0;
  REP(i, n) sum += a[i];
  if (sum % (n * (n + 1) / 2) != 0) {
    no();
  }
  auto idx = [&](ll i) { return (i + n) % n; };
  ll k = sum / (n * (n + 1) / 2);
  REP(i, n) {
    ll d = k + a[idx(i)] - a[idx(i + 1)];
    if (d % n != 0) no();
    ll t = d / n;
    if (t < 0 || k < t) no();
  }
  yes();
}