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;
/*{{{*/  //template
#define rep(i,n) for(int i=0;i<(int)(n);i++)
#define rrep(i,n) for(int i=(n)-1;i>=0;i--)
constexpr int INF = numeric_limits<int>::max()/2;
constexpr long long LINF = numeric_limits<long long>::max()/3;
#define pb push_back
#define eb emplace_back
#define fi first
#define se second
#define all(v) (v).begin(),(v).end()
#define sz(x) (int)(x).size()
#define debug(x) cerr<<#x<<":"<<x<<endl
#define debug2(x,y) cerr<<#x<<","<<#y":"<<x<<","<<y<<endl
//struct fin{ fin(){ cin.tie(0); ios::sync_with_stdio(false); } } fin_;
struct Double{ double d; explicit Double(double x) : d(x){} };
ostream& operator<<(ostream& os,const Double x){ os << fixed << setprecision(20) << x.d; return os; }
template<typename T> ostream& operator<<(ostream& os,const vector<T>& vec){ os << "["; for(const auto& v : vec){ os << v << ","; } os << "]"; return os; }
template<typename T,typename U> ostream& operator<<(ostream& os, const map<T,U>& mp){ os << "{"; for(auto& p : mp){ os << p << ","; } os << "}"; return os; }
template<typename T,typename U> ostream& operator<<(ostream& os,const pair<T,U>& p){ os << "(" << p.first << ","<< p.second <<")"; return os; }
template<typename T> ostream& operator<<(ostream& os,const set<T>& st){ os<<"{"; for(T v:st) os<<v<<","; os <<"}"; return os; }
template<typename T,typename U> inline void chmax(T &x,U y){ if(y>x) x = y; }
template<typename T,typename U> inline void chmin(T &x,U y){ if(y<x) x = y; }
using ll = long long;
using ull = unsigned long long;
using pii = std::pair<int,int>;
using vi = std::vector<int>;
typedef vector<vi> vvi;
ll gcd(ll a,ll b){ if(b==0) return a; else return gcd(b,a%b); }
constexpr ll TEN(int n) { return (n == 0) ? 1 : 10 * TEN(n - 1); }
//constexpr double eps = 1e-14; 
constexpr double eps = 1e-10; 
constexpr ll mod = 1e9+7;
const int dx[]={1,0,-1,0} ,dy[] = {0,1,0,-1};
/*}}}*/

int main(){
    ll N; cin >> N;
    vector<ll> a(N+1);
    rep(i,N) cin >> a[i];
    a[N] = a[0];

    vector<ll> b(N);
    rep(i,N) b[i] = a[i+1] - a[i];

    ll target = *max_element(all(b));
    vector<ll> c(N);
    rep(i,N){
        if((target - b[i])%N){
            cout << "NO" << endl;
            return 0;
        }
        c[i] = (target - b[i]) / N;
    }


    ll p = a[0];
    ll p2 = a[1];
    rep(i,N){
        p -= (N-i) * c[i];

        ll t = (N-i+1)%N;
        if(t==0) t=N;
        p2 -= c[i] * t;
    }
    debug(p);
    debug(p2);

    ll sum = N * (N+1) / 2;
    if(p<0 || (p%sum>0)){
        cout << "NO" << endl;
        return 0;
    }

    cout << "YES" << endl;
}