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;

#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define FOR(i, a, b) for(int i=(a);i<(b);i++)
#define REP(i, n) FOR(i, 0, n)
#define RFOR(i, a, b) for(int i=(a);i>=(b);i--)
#define RREP(i, n) RFOR(i, n, 0)
#define MFOR(i, m) for(auto i=(m).begin();i!=(m).end();i++)
#define ALL(a) (a).begin(), (a).end()
#define SZ(x) ((int)(x).size())

typedef long long int ll;
typedef pair<int, int> P;
typedef pair<ll, ll> Pll;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<ll> vll;
typedef vector<vll> vvll;

const double eps = 1e-10;
const int MOD = 1000000007;
const int INF = 1000000000;
const ll LINF = 1 << 30;

template<typename T>
void printv(vector<T> const& s) {
  REP(i, SZ(s)) {
    cout << s[i] << " ";
  }
  cout << endl;
}

int main () {
  cin.tie(0);
  cout << setprecision(10);

  ll n; cin >> n;
  vll a(n+1);
  ll s = 0;
  REP(i, n) {
    cin >> a[i];
    s += a[i];
  }
  a[n] = a[0];
  vll b(n);
  REP(i, n) {
    b[i] = a[i+1] - a[i];
  }

  // printv(b);


  ll su = n * (n + 1) / 2ll;
  // cout << s << ":" << su << endl;
  if(s % su == 0) {
    ll cnt = 0;
    REP(i, n) {
      b[i] -= s / su;
      if(abs(b[i]) % n != 0) {
        cout << "NO" << endl;
        return 0;
      }
      cnt += abs(b[i]) / n;
    }
    if(cnt == s / su) cout << "YES" << endl;
    else cout << "NO" << endl;
  } else {
    cout << "NO" << endl;
  }
}