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>
#include <ext/hash_map>
#include <ext/numeric>

using namespace std;
using namespace __gnu_cxx;

#define REP(i,n) for( (i)=0 ; (i)<(n) ; (i)++ )
#define rep(i,x,n) for( (i)=(x) ; (i)<(n) ; (i)++ )
#define REV(i,n) for( (i)=(n) ; (i)>=0 ; (i)-- )
#define FORIT(it,x) for( (it)=(x).begin() ; (it)!=(x).end() ; (it)++ )
#define foreach(it,c) for(__typeof((c).begin()) it=(c).begin();it!=(c).end();++it)
#define rforeach(it,c) for(__typeof((c).rbegin()) it=(c).rbegin();it!=(c).rend();++it)
#define foreach2d(i, j, v) foreach(i,v) foreach(j,*i)
#define all(x) (x).begin(),(x).end()
#define rall(x) (x).rbegin(),(x).rend()
#define SZ(x) ((int)(x).size())
#define MMS(x,n) memset(x,n,sizeof(x))
#define mms(x,n,s) memset(x,n,sizeof(x)*s)
#define pb push_back
#define mp make_pair
#define NX next_permutation
#define UN(x) sort(all(x)),x.erase(unique(all(x)),x.end())
#define CV(x,n) count(all(x),(n))
#define FIND(x,n) find(all(x),(n))-(x).begin()
#define ACC(x) accumulate(all(x),0)
#define PPC(x) __builtin_popcountll(x)
#define LZ(x) __builtin_clz(x)
#define TZ(x) __builtin_ctz(x)
#define mxe(x) *max_element(all(x))
#define mne(x) *min_element(all(x))
#define low(x,i) lower_bound(all(x),i)
#define upp(x,i) upper_bound(all(x),i)
#define NXPOW2(x) (1ll << ((int)log2(x)+1))
#define PR(x) cout << #x << " = " << (x) << endl ;

typedef unsigned long long ull;
typedef long long ll;
typedef vector<int> vi;
typedef vector<vector<int> > vvi;
typedef pair<int, int> pii;

const int OO = (int) 2e9;
const double eps = 1e-9;

const int N = 100005;

int n;
int a[N], d[N];
ll sum, tot;

int main() {
	std::ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	cout.tie(NULL);
#ifndef ONLINE_JUDGE
//	freopen("in.txt", "rt", stdin);
//	freopen("out.txt", "wt", stdout);
#endif
	cin >> n;
	for (int i = 0; i < n; i++) {
		cin >> a[i];
		sum += a[i];
	}
	tot = n * (n + 1LL) / 2;
	if (sum % tot) {
		puts("NO");
		return 0;
	}
	ll steps = sum / tot;
	ll s = 0;
	for (int i = 0; i < n - 1; i++) {
		d[i] = (a[i + 1] - a[i]);
		if ((steps - d[i]) < 0 || (steps - d[i]) % n) {
			puts("NO");
			return 0;
		}
		s += (steps - d[i]) / n;
	}
	puts("YES");
	return 0;
}