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 ADD(a, b) a = (a + ll(b)) % mod
#define MUL(a, b) a = (a * ll(b)) % mod
#define MAX(a, b) a = max(a, b)
#define MIN(a, b) a = min(a, b)
#define rep(i, a, b) for(int i = int(a); i < int(b); i++)
#define rer(i, a, b) for(int i = int(a) - 1; i >= int(b); i--)
#define all(a) (a).begin(), (a).end()
#define sz(v) (int)(v).size()
#define pb push_back
#define sec second
#define fst first
#define debug(fmt, ...) Debug(__LINE__, ":", fmt, ##__VA_ARGS__)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pi;
typedef pair<ll, ll> pl;
typedef pair<int, pi> ppi;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<vl> mat;
typedef complex<double> comp;
void Debug() {cout << '\n'; }
template<class FIRST, class... REST>void Debug(FIRST arg, REST... rest){
	cout<<arg<<" ";Debug(rest...);}
template<class T>ostream& operator<<(ostream& out,const vector<T>& v) {
	out<<"[";if(!v.empty()){rep(i,0,sz(v)-1)out<<v[i]<<", ";out<<v.back();}out<<"]";return out;}
template<class S, class T>ostream& operator<<(ostream& out,const pair<S, T>& v){
	out<<"("<<v.first<<", "<<v.second<<")";return out;}
const int MAX_N = 200010;
const int MAX_V = 100010;
const double eps = 1e-6;
const ll mod = 1000000007;
const int inf = 1 << 29;
const ll linf = 1LL << 60;
const double PI = 3.14159265358979323846;
///////////////////////////////////////////////////////////////////////////////////////////////////

int N;
ll A[MAX_N];

void solve() {
	cin >> N;
	ll S = 0;
	rep(i, 0, N) {
		cin >> A[i];
		S += A[i];
	}
	ll D = N * (N + 1) / 2;
	if(S % D != 0) {
		cout << "NO\n"; return;
	}
	else {
		ll T = S / D;
		rep(i, 0, N) {
			int to = (i + 1) % N;
			if(A[to] % N != (T + A[i]) % N) {
				cout << "NO\n"; return;
			}
		}
	}
	cout << "YES\n";
}

int main() {
#ifndef LOCAL
	ios::sync_with_stdio(false);
    cin.tie(0);
#endif
    cout << fixed;
	cout.precision(20);
	srand((unsigned int)time(NULL));
#ifdef LOCAL
	//freopen("in.txt", "wt", stdout); //for tester
    freopen("in.txt", "rt", stdin);
#endif	
	solve();
#ifdef LOCAL
    cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.\n";
#endif
	return 0;
}