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)

#ifdef LOCAL
//#define _GLIBCXX_DEBUG
#endif
//#pragma GCC target("avx512f,avx512dq,avx512cd,avx512bw,avx512vl")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef pair<ll, ll> P;
typedef pair<int, int> Pi;
typedef vector<ll> Vec;
typedef vector<int> Vi;
typedef vector<string> Vs;
typedef vector<P> VP;
typedef vector<vector<ll>> VV;
typedef vector<vector<int>> VVi;
typedef vector<vector<vector<ll>>> VVV;
typedef vector<vector<vector<vector<ll>>>> VVVV;

#define REP(i, a, b) for(ll i=(a); i<(b); i++)
#define PER(i, a, b) for(ll i=(a); i>=(b); i--)
#define rep(i, n) REP(i, 0, n)
#define per(i, n) PER(i, n, 0)
const ll INF=1e18+18;
const ll MOD=1000000007;
#define Yes(n) cout << ((n) ? "Yes" : "No") << endl;
#define YES(n) cout << ((n) ? "YES" : "NO") << endl;
#define ALL(v) v.begin(), v.end()
#define rALL(v) v.rbegin(), v.rend()
#define pb(x) push_back(x)
#define mp(a, b) make_pair(a,b)
#define Each(a,b) for(auto &a :b)
#define REPM(i, mp) for (auto i = mp.begin(); i != mp.end(); ++i)
#ifdef LOCAL
#define dbg(x_) cerr << #x_ << ":" << x_ << endl;
#define dbgmap(mp) cerr << #mp << ":"<<endl; for (auto i = mp.begin(); i != mp.end(); ++i) { cerr << i->first <<":"<<i->second << endl;}
#define dbgarr(n,m,arr) rep(i,n){rep(j,m){cerr<<arr[i][j]<<" ";}cerr<<endl;}
#define dbgdp(n,arr) rep(i,n){cerr<<arr[i]<<" ";}cerr<<endl;
#define dbgmint(n,arr) rep(i,n){cerr<<arr[i].x<<" ";}cerr<<endl;
#define dbgarrmint(n,m,arr) rep(i,n){rep(j,m){cerr<<arr[i][j].x<<" ";}cerr<<endl;}
#else
#define dbg(...)
#define dbgmap(...)
#define dbgarr(...)
#define dbgdp(...)
#define dbgmint(...)
#define dbgarrmint(...)
#endif
#define out(a) cout<<a<<endl
#define out2(a,b) cout<<a<<" "<<b<<endl
#define vout(v) rep(i,v.size()){cout<<v[i]<<" ";}cout<<endl
#define Uniq(v) v.erase(unique(v.begin(), v.end()), v.end())
#define fi first
#define se second

template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return true; } return false; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return true; } return false; }

template<typename T1, typename T2>
ostream &operator<<(ostream &s, const pair<T1, T2> &p) { return s<<"("<<p.first<<", "<<p.second<<")"; }

template<typename T>istream& operator>>(istream&i,vector<T>&v)
{rep(j,v.size())i>>v[j];return i;}

// vector
template<typename T>
ostream &operator<<(ostream &s, const vector<T> &v) {
    int len=v.size();
    for(int i=0; i<len; ++i) {
        s<<v[i];
        if(i<len-1) s<<"	";
    }
    return s;
}

// 2 dimentional vector
template<typename T>
ostream &operator<<(ostream &s, const vector<vector<T> > &vv) {
    int len=vv.size();
    for(int i=0; i<len; ++i) {
        s<<vv[i]<<endl;
    }
    return s;
}

//mint
struct mint {
    ll x; // typedef long long ll;
    mint(ll x=0):x((x%MOD+MOD)%MOD){}
    mint operator-() const { return mint(-x);}
    mint& operator+=(const mint a) {
        if ((x += a.x) >= MOD) x -= MOD;
        return *this;
    }
    mint& operator-=(const mint a) {
        if ((x += MOD-a.x) >= MOD) x -= MOD;
        return *this;
    }
    mint& operator*=(const mint a) { (x *= a.x) %= MOD; return *this;}
    mint operator+(const mint a) const { return mint(*this) += a;}
    mint operator-(const mint a) const { return mint(*this) -= a;}
    mint operator*(const mint a) const { return mint(*this) *= a;}
    mint pow(ll t) const {
        if (!t) return 1;
        mint a = pow(t>>1);
        a *= a;
        if (t&1) a *= *this;
        return a;
    }

    // for prime MOD
    mint inv() const { return pow(MOD-2);}
    mint& operator/=(const mint a) { return *this *= a.inv();}
    mint operator/(const mint a) const { return mint(*this) /= a;}
};
istream& operator>>(istream& is, const mint& a) { return is >> a.x;}
ostream& operator<<(ostream& os, const mint& a) { return os << a.x;}
typedef vector<mint> Vmint;

struct combination {
    vector<mint> fact, ifact;
    combination(int n):fact(n+1),ifact(n+1) {
        assert(n < MOD);
        fact[0] = 1;
        for (int i = 1; i <= n; ++i) fact[i] = fact[i-1]*i;
        ifact[n] = fact[n].inv();
        for (int i = n; i >= 1; --i) ifact[i-1] = ifact[i]*i;
    }
    mint operator()(int n, int k) {
        if (k < 0 || k > n) return 0;
        return fact[n]*ifact[k]*ifact[n-k];
    }
};

void simulate(){

    ll n = 5;
    Vec v(n);
    ll t = 3;

    random_device rd;
    mt19937 mt(rd());
    rep(i,t){
        ll r = mt()%n;
        rep(k,n){
            v[k] += (k+r)%n+1;
        }
    }
    ll sum = accumulate(ALL(v),0);
    ll step = (n+1)*n/2;
    ll cnt = sum/step;

    dbg(v);

    ll sum2 = 0;
    rep(i,n){
        v[i] -= cnt*i;
        sum2 += v[i];
    }
    assert(sum2 == cnt * n);
    dbg(v);
    exit(0);
}

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    cout<<std::setprecision(10);

    //simulate();
    ll n;
    cin>>n;
    Vec a(n);
    cin>>a;
    ll sum = 0;
    rep(i,n)sum+=a[i];
    ll step = (n+1)*n/2;
    ll cnt = sum/step;

    bool ok = true;
    if(sum%step!=0) ok = false;

    ll sum2 = 0;
    rep(i,n){
        a[i] -= cnt*i;
        sum2 += a[i];
    }
    if(sum2 != n*cnt) ok = false;
    REP(i,1,n){
        ll diff = a[i]-a[i-1];
        if(diff%n!=0)ok = false;
    }
    YES(ok);

    return  0;
}