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: Runtime Error)

//Zory-2020
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
// typedef __int128 ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
#define FR first
#define SE second
#define MP make_pair
#define PB push_back
#define vc vector
#define db double
#define all(x) (x).begin(),(x).end()
#define sz(x) ((int)(x).size())
#define bin(x) (1ll<<(x))
#define fo(i,l,r) for(int i=(l),I=(r);i<=I;i++)
#define fd(i,r,l) for(int i=(r),I=(l);i>=I;i--)
#define mem(x,val) memset(x,val,sizeof x)
#define Swap(a,b,n) for(int I=0;I<=n;I++) swap(a[I],b[I])
#define PC __builtin_popcountll
#ifdef DEBUG
	#define debug(A,args...) fprintf(stderr,A,##args)
#else
	#define debug(A,args...) 
#endif
namespace mine
{
	ll qread()
	{
		ll ans=0,f=1;char c=getchar();
		while(c<'0' or c>'9') {if(c=='-')f=-1;c=getchar();}
		while('0'<=c and c<='9') ans=ans*10+c-'0',c=getchar();
		return ans*f;
	}
	void write(ll num){if(num<0) putchar('-'),num=-num;if(num>=10) write(num/10);putchar('0'+num%10);}
	void write1(ll num){write(num);putchar(' ');}
	void write2(ll num){write(num);putchar('\n');}
	template<typename T>inline bool chmax(T&a,const T&b){return a<b?a=b,1:0;}
	template<typename T>inline bool chmin(T&a,const T&b){return a>b?a=b,1:0;}
	ll gcd(ll x,ll y){return y?gcd(y,x%y):x;}
	bool IN(ll x,ll l,ll r){return l<=x and x<=r;}
	void GG(){puts("NO");exit(0);}
	
	const db eps=1e-8;
	const int INF=0x3f3f3f3f;
	const int MOD=1e9+7;
	int mm(const int x){return x>=MOD?x-MOD:x;}
	template<typename T> void add(T &x,const int &y){x=(x+y>=MOD?x+y-MOD:x+y);}
	ll qpower(ll x,ll e,int mod=MOD){ll ans=1;while(e){if(e&1)ans=ans*x%mod;x=x*x%mod;e>>=1;}return ans;}
	ll invm(ll x){return qpower(x,MOD-2);}
	const int M=5e6+10;
	ll fac[M],facinv[M],Inv[M];ll C(int n,int m){return n<0 or n<m?0:fac[n]*facinv[m]%MOD*facinv[n-m]%MOD;}
	void PRE()
	{
		fac[0]=1;fo(i,1,M-1) fac[i]=fac[i-1]*i%MOD;
		facinv[M-1]=invm(fac[M-1]);fd(i,M-1,1) facinv[i-1]=facinv[i]*i%MOD;
		Inv[1]=1;fo(i,2,M-1) Inv[i]=(MOD-MOD/i)*Inv[MOD%i]%MOD;
	}
	const int N=4e3+10;
	//------------------FIXED------------------
	double a[N][N];
	void main()
	{
		int n=qread();if(n>500) exit(1);
		fo(i,1,n){ fo(j,1,n) a[i][(i-j<1?i+n-j:i-j)]=j;a[i][0]=qread(); }
		fo(u,1,n)
		{
			int wt=0;fo(i,1,n) if(fabs(a[i][u])>eps) wt=i;if(!wt) GG();
			fo(t,0,n) swap(a[u][t],a[wt][t]);
			fo(j,1,n) if(j!=u and fabs(a[j][u])>eps)
			{
				double tt=a[j][u]/a[u][u];
				fo(t,0,n) a[j][t]-=tt*a[u][t];
			}
		}
		fo(u,1,n) if(fabs(a[u][0]/a[u][u]-round(a[u][0]/a[u][u]))>eps) GG();puts("YES");
	}
};//变量重名！
signed main()
{
	#ifdef DEBUG
	//freopen("a.in","r",stdin);
	freopen("z.txt","r",stdin);
	//freopen("a.out","w",stdout);
	#endif
	srand(time(0));
	mine::PRE();//线性预处理模意义
	mine::main();
	debug("\n---------------------Zory---------------------\nTime: %.2lf s",1.0*clock()/CLOCKS_PER_SEC);
}