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)

// tested by Hightail: https://github.com/dj3500/hightail
#include <iostream>
#include <cstdio>
#include <string>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <cmath>
#include <algorithm>
#include <sstream>
#include <stack>
#include <cstring>
#include <iomanip>
#include <ctime>
#include <cassert>
using namespace std;
#define pb push_back
#define INF 1001001001
#define FOR(i,n) for(int (i)=0;(i)<(n);++(i))
#define FORI(i,n) for(int (i)=1;(i)<=(n);++(i))
#define mp make_pair
#define pii pair<int,int>
#define ll long long
#define vi vector<int>
#define SZ(x) ((int)((x).size()))
#define fi first
#define se second
#define wez(n) int (n); scanf("%d",&(n));
#define wez2(n,m) int (n),(m); scanf("%d %d",&(n),&(m));
#define wez3(n,m,k) int (n),(m),(k); scanf("%d %d %d",&(n),&(m),&(k));
inline void pisz(int n) { printf("%d\n",n); }
template<typename T,typename TT> ostream& operator<<(ostream &s,pair<T,TT> t) {return s<<"("<<t.first<<","<<t.second<<")";}
template<typename T> ostream& operator<<(ostream &s,vector<T> t){FOR(i,SZ(t))s<<t[i]<<" ";return s; }
#define DBG(vari) cout<<"["<<__LINE__<<"] "<<#vari<<" = "<<(vari)<<endl;
#define ALL(t) t.begin(),t.end()
#define FOREACH(i,t) for (__typeof(t.begin()) i=t.begin(); i!=t.end(); i++)
#define TESTS wez(testow)while(testow--)
#define REP(i,a,b) for(int (i)=(a);(i)<=(b);++i)
#define REPD(i,a,b) for(int (i)=(a); (i)>=(b);--i)
#define REMAX(a,b) (a)=max((a),(b));
#define REMIN(a,b) (a)=min((a),(b));
#define IOS ios_base::sync_with_stdio(0);

const int N = 200200;
ll n, a[N];

void fail () {
   printf("NO");
   exit(0);
}

int main () {
   IOS
   cin >> n;
   FOR(i,n) cin >> a[i];
   ll sum = 0;
   FOR(i,n) sum += a[i];
   if (sum % (n*(n+1)/2) != 0) fail();
   ll k = sum / (n*(n+1)/2);
   ll sk = 0;
   FOR(i,n) {
      ll t = (a[i] + k) - a[(i+1)%n];
      if (t < 0 || t % n != 0) fail();
      sk += t/n;
   }
   if (k != sk) fail();
   printf("YES");
}