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)

#include <bits/stdc++.h>

using namespace std;

using ll = long long;
using dbl = double;
using pii = pair<int, int>;
using pl4 = pair<ll, ll>;
using vi = vector<int>;
using vvi = vector<vi>;
using vs = vector<string>;
using vvs = vector<vs>;
using vll = vector<ll>;
using vvll = vector<vll>;
using vvvll = vector<vvll>;
using vpii = vector<pii>;
using vvpii = vector<vpii>;
using vpl4 = vector<pl4>;
using vvpl4 = vector<vpl4>;
using vd = vector<dbl>;
using vvd = vector<vd>;

#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define pob pop_back()
#define sz size()
#define be begin()
#define en end()
#define asn assign
#define emp empty()
#define ft front()
#define bk back()
#define clr clear()
#define ins insert
#define ers erase
#define res resize
#define tp top()

#define FOR(i,a,b) for(int i=(a);i<=(b);i++)
#define rFOR(i,a,b) for(int i=(b);i>=(a);i--)
#define ROR(v,i) for(auto &(i):(v))
#define SORT(a) sort((a).be,(a).en)
#define rSORT(a) sort((a).rbegin(),(a).rend())
#define UNIQUE(a) (a).erase(unique((a).be,(a).en),(a).en)
#define PREVP(a) prev_permutation((a).be,(a).en)
#define NEXTP(a) next_permutation((a).be,(a).en)
#define BINS(a,b) binary_search((a).be,(a).en,(b))
#define LOWB(a,b) (lower_bound((a).be,(a).en,(b))-(a).be)
#define UPB(a,b) (upper_bound((a).be,(a).en,(b))-(a).be)
#define CNT(a,b) count((a).be,(a).en,b)
#define SUM(a) accumulate((a).be,(a).en,0)
#define REV(a) reverse((a).be,(a).en)
#define REGS(a,b) regex_search((a),regex(b))
#define REGM(a,b) regex_match((a),regex(b))
#define yn(a) cout <<((a)?"yes":"no")<<endl;
#define Yn(a) cout <<((a)?"Yes":"No")<<endl;
#define YN(a) cout <<((a)?"YES":"NO")<<endl;
#define say(a) cout <<(a);
#define sal(a) cout <<(a)<<endl;
#define sas cout <<" ";
#define sak cout <<endl;
#define dbg(a) cout <<(#a)<<": "<<(a)<<endl;
#define c2l(a) ((ll)(a-48))
#define a2l(a) ((ll)(a-97))
#define A2l(a) ((ll)(a-65))
#define l2c(a) ((char)(a+48))
#define l2a(a) ((char)(a+97))
#define l2A(a) ((char)(a+65))
#define DigN2(a) ((llabs(a)==0)?(1):((ll)(log2(double(llabs(a))))+1))
#define DigN10(a) ((llabs(a)==0)?(1):((ll)(log10(double(llabs(a))))+1))
#define Dig2(a,b) (((a)>>(b))&1)
#define Dig10(a,b) (ll)(((a)/((ll)(pow(10.0,(double)(b)))))%10)
#define Pow2(a) (1<<(a))
#define Pow10(a) ((ll)(pow(10.0,double(a))))
#define llin(a) ll (a);cin >>(a);
#define stin(a) string (a);cin >>(a);
#define vin(v) ROR((v),(i)){cin >>(i);};
#define vllin(v,N) vll (v)((N));vin(v);
#define vsin(v,N) vs (v)((N));vin(v);
#define rdn(a,b) ((a)/(b))
#define rou(a,b) ((((double(a)/double(b))-((a)/(b)))<0.5)?((a)/(b)):(((a)/(b))+1))
#define rup(a,b) ((((a)%(b))==0)?((a)/(b)):(((a)/(b))+1))
#define min(a,b) ((a<b)?(a):(b))
#define max(a,b) ((a>b)?(a):(b))
#define powll(a,b) (ll)(pow((double)(a),(double)(b)))
#define Triangle(x1,y1,x2,y2,x3,y3) (((x1)-(x2))*((y1)-(y3))-((x1)-(x3))*((y1)-(y2)))

#define int ll

const ll MOD = 1e9 + 7;
//const ll MOD = 998244353;
//const ll MOD = 9007199254740881;
const ll INF = 1LL << 60;
const string alp = "abcdefghijklmnopqrstuvwxyz";
const string ALP = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";

ll gcd(ll a, ll b) { if (b == 0)return a; return gcd(b, a%b); }
ll lcm(ll a, ll b) { return a / gcd(a, b)*b; }

pl4 Bezout(ll a, ll b) {
	if (b != 0) {
		pl4 xy;
		xy = Bezout(b, a%b);
		return mp(xy.se, xy.fi - ((a / b)*xy.se));
	}
	if (b == 0) {
		return mp(1, 0);
	}
}
pl4 Bez(ll a, ll b, ll c) {
	pl4 xy;
	ll x, y, z, gc;
	xy = Bezout(a, b);
	gc = gcd(a, b);
	if (c%gc != 0) return mp(-1, -1);
	x = xy.fi*(c / gc); y = xy.se*(c / gc);
	if (x < 0) z = rup(-x, (b / gc));
	if (x >= 0) z = -x / (b / gc);
	x += z * (b / gc);
	y -= z * (a / gc);
	return mp(x, y);
}

void salv(vll v) {
	say("{");
	FOR(i, 0, v.sz - 1) {
		say(v[i]);
		if (i != v.sz - 1) say(",");
	}
	sal("}")
}

ll DigS10(ll n) {
	ll m = 0;
	FOR(i, 0, DigN10(n) - 1) {
		m += (ll)((llabs(n) % (ll)(pow(10.0, (double)(i + 1)))) / (ll)(pow(10.0, (double)i)));
	}
	return m;
}

ll isP(ll n) {
	if (n <= 1) return 0;
	FOR(i, 2, (ll)sqrt(n)) {
		if (n%i == 0) return 0;
	}
	return 1;
}

vll FactM(1, 1);
vll FactMI(1, 1);

ll PowM(ll a, ll b) {
	ll ans = 1, x = (a%MOD);
	FOR(i, 0, DigN2(b) - 1) {
		if (Dig2(b, i) == 1) ans = (ans*x) % MOD;
		if (i != (DigN2(b) - 1)) x = (x*x) % MOD;
	}
	return ans;
}

void CFactM(ll n) {
	if (FactM.sz <= n) {
		FOR(i, FactM.sz, n) {
			FactM.pb((FactM[i - 1] * (i%MOD)) % MOD);
		}
	}
	return;
}

void CFactMI(ll n) {
	CFactM(n);
	if (FactMI.sz < (n + 1)) FactMI.res(n + 1, -1);
	if (FactMI[n] == -1) FactMI[n] = PowM(FactM[n], MOD - 2);
	rFOR(i, 1, n - 1) {
		if (FactMI[i] != -1) break;
		FactMI[i] = ((FactMI[i + 1] * ((i + 1) % MOD)) % MOD);
	}
	return;
}

ll CombM(ll n, ll k) {
	if ((n < 0) || (k < 0)) return 0;
	if (n < k) return 0;
	if (n + 1 > FactMI.sz) CFactMI(n);
	return ((((FactMI[k] * FactMI[n - k]) % MOD)*FactM[n]) % MOD);
}

signed main() {

	llin(N);
	vllin(A, N);
	ll ca = 0;
	ROR(A, i) ca += i;
	ll can = 1;
	if (ca % (((N + 1)*N) / 2) != 0) can = 0;
	else {
		ROR(A, i) i %= N;
		ll aa = (A[1] - A[0] + N) % N;
		FOR(i, 1, N - 1) {
			if ((A[(i + 1) % N] - A[i] + N) % N != aa) can = 0;
		}
	}
	YN(can);

}