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 <iostream>
#include <cstdio>
#include <vector>
#define _USE_MATH_DEFINES
#include <math.h>
#include <cstring>
#include <numeric>
#include <algorithm>
#include <stdlib.h>
#include <functional>
#include <string>
#include <list>
#include <fstream>
#include <iomanip>
#include <array>
#include <map>
#include <queue>
#include <limits.h>
#include <set>
#include <stack>
#include <random>
#include <complex>
#include <unordered_map>
#include <unordered_set>
#include <nmmintrin.h>
#include <chrono>
#define rep(i,s,n) for(int i = (s); (n) > i; i++)
#define REP(i,n) rep(i,0,n)
#define RANGE(x,a,b) ((a) <= (x) && (x) <= (b))
#define DUPLE(a,b,c,d) (RANGE(a,c,d) || RANGE(b,c,d) || RANGE(c,a,b) || RANGE(d,a,b))
#define INCLU(a,b,c,d) (RANGE(a,c,d) && (b,c,d))
#define PW(x) ((x)*(x))
#define ALL(x) (x).begin(), (x).end()
#define MODU 1000000007
#define bitcheck(a,b)   ((a >> b) & 1)
#define bitset(a,b)      ( a |= (1 << b))
#define bitunset(a,b)    (a &= ~(1 << b))
#define MP(a,b) make_pair((a),(b))
#define Manh(a,b) (abs((a).first-(b).first) + abs((a).second - ((b).second))
#define pritnf printf
#define scnaf scanf
#define itn int
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
ll gcd(ll a, ll b) {
	if (b == 0) return a;
	return gcd(b, a%b);
}
template<typename A, size_t N, typename T>
void Fill(A(&array)[N], const T &val) {
	std::fill((T*)array, (T*)(array + N), val);
}

struct Edge {
	int from, to;
};

signed main()
{
	ll n;

	cin >> n;
	vector<ll> st(n),sa(n-1);
	ll sum = 0, su = (1 + n) * n / 2;
	REP(i, n) {
		scanf("%lld", &st[i]);
		sum += st[i];

		if (i)
			sa[i - 1] = st[i] - st[i - 1];
	}
	if(sum % su) {
		cout << "NO" << endl;
		return 0;
	}
	ll t = sum / su;
	ll rt = t;
	ll cs = 0;
	REP(i, n-1) {
		if ((t - sa[i]) % n) {
			cout << "NO" << endl;
			return 0;
		}
		
		rt -= (t - sa[i]) / n;
	}
	if(rt < 0) {
		cout << "NO" << endl;
		return 0;
	}

	cout << "YES" << endl;
	return 0;
}