Problem:
Score : 300 points

Problem Statement

Snuke has N hats. The i-th hat has an integer a_i written on it.

There are N camels standing in a circle.
Snuke will put one of his hats on each of these camels.

If there exists a way to distribute the hats to the camels such that the following condition is satisfied for every camel, print Yes; otherwise, print No.

The bitwise XOR of the numbers written on the hats on both adjacent camels is equal to the number on the hat on itself.

What is XOR?

The bitwise XOR x_1 \oplus x_2 \oplus \ldots \oplus x_n of n non-negative integers x_1, x_2, \ldots, x_n is defined as follows:

- When x_1 \oplus x_2 \oplus \ldots \oplus x_n is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if the number of integers among x_1, x_2, \ldots, x_n whose binary representations have 1 in the 2^k's place is odd, and 0 if that count is even.

For example, 3 \oplus 5 = 6.

Constraints

All values in input are integers.

3 \leq N \leq 10^{5}

0 \leq a_i \leq 10^{9}

Input

Input is given from Standard Input in the following format:

N
a_1 a_2 \ldots a_{N}

Output

Print the answer.

Sample Input 1

3
1 2 3

Sample Output 1

Yes

If we put the hats with 1, 2, and 3 in this order, clockwise, the condition will be satisfied for every camel, so the answer is Yes.

Sample Input 2

4
1 2 4 8

Sample Output 2

No

There is no such way to distribute the hats; the answer is No.

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#pragma GCC target("avx")
//#pragma GCC target("avx512f,avx512dq,avx512cd,avx512bw,avx512vl")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#include <bits/stdc++.h>
using namespace std;
#define LL long long
#define DB double
#define LD long double
#define ST string
#define BS bitset
#define PA pair<LL,LL>
#define VE vector
#define VL vector<LL>
#define VP vector<pair<LL,LL>>
#define VVL vector<vector<LL>>
#define PQ priority_queue
#define PQS priority_queue<LL,vector<LL>,greater<LL>>
#define PB push_back
#define POB pop_back
#define PF push_front
#define POF pop_front
#define EB emplace_back
#define MP make_pair
#define TS to_string
#define TU to_ullong
#define BPL __builtin_popcountll
#define FOR(i,a,n) for(i=a;i<n;++i)
#define FORR(i,a,n) for(i=n-1;i>=a;--i)
#define rep(i,n) FOR(i,0,n)
#define repr(i,n) FORR(i,0,n)
#define ALL(a) a.begin(),a.end()
#define RALL(a) a.rbegin(),a.rend()
#define SORT(a) sort(ALL(a))
#define REV(a) reverse(ALL(a))
#define UB(a,n) *upper_bound(ALL(a),n)
#define LB(a,n) *lower_bound(ALL(a),n)
#define INF 1145141919810364364
#define PI 3.14159265358979
//#define MOD 1000000007
#define MOD 998244353
#define ERR 0.00000001
#define FAST cin.tie(0);ios::sync_with_stdio(false)

void Yn(LL a){if(a)printf("Yes\n");else printf("No\n");}
void YN(LL a){if(a)printf("YES\n");else printf("NO\n");}
LL pwmn(LL a,LL n){LL ans=1;while(ans<a)ans*=n;return ans;}
LL dig(LL n){LL ret=0;while(n)n/=10,ret++;return ret;}
LL LBn(VL &v,LL a){LL mx=v.size(),mn=-1,md;while(mx-mn>1){md=(mx+mn)/2;if(v[md]<a)mn=md;else mx=md;}return mx;}
LL GCD(LL a,LL b){LL c=1,tmp=max(a,b);b=min(a,b);a=tmp;while(c!=0){c=a%b;a=b;b=c;}return a;}
LL LCM(LL a,LL b){return a*b/GCD(a,b);}
int lcmp(const void* a,const void* b){if(*(LL*)a>*(LL*)b)return 1;if(*(LL*)a<*(LL*)b)return -1;return 0;}
int lcmpr(const void* a,const void* b){if(*(LL*)a>*(LL*)b)return -1;if(*(LL*)a<*(LL*)b)return 1;return 0;}
int ccmp(const void* a,const void* b){return *(char*)a-*(char*)b;}
int ccmpr(const void* a,const void* b){return *(char*)b-*(char*)a;}
int scmp(const void* a,const void* b){return strcmp((char*)a,(char*)b);}
int scmpr(const void* a,const void* b){return strcmp((char*)b,(char*)a);}
LL cmod(LL a,LL m){if(a%m<0)return a%m+abs(m);else return a%m;}
LL DIV(LL a,LL d,LL m){LL l=m,x=1,y=0,k;while(l){k=d/l;d-=k*l;swap(l,d);x-=k*y;swap(x,y);}return cmod(a*cmod(x,m),m);}
LL POW(LL a,LL n,LL m){LL ans=1;while(n>0){if(n&1)ans=ans*a%m;a=a*a%m;n>>=1;}return ans;}
VL fact,finv,inv;
void comi(LL n){LL i;fact.resize(max(2LL,n+1));finv.resize(max(2LL,n+1));inv.resize(max(2LL,n+1));fact[0]=fact[1]=1;finv[0]=finv[1]=1;inv[1]=1;FOR(i,2,n+1){fact[i]=fact[i-1]*i%MOD;inv[i]=MOD-inv[MOD%i]*(MOD/i)%MOD;finv[i]=finv[i-1]*inv[i]%MOD;}}
LL com(LL n,LL k){if(n<k||n<0||k<0)return 0;return fact[n]*(finv[k]*finv[n-k]%MOD)%MOD;}
bool cmps(pair<LL,LL> a,pair<LL,LL> b){if(a.second!=b.second)return a.second<b.second;return a.first<b.first;}
template<typename T>bool chmax(T &a,T b){if(a<b){a=b;return true;}return false;}
template<typename T>bool chmin(T &a,T b){if(a>b){a=b;return true;}return false;}

int main(){
	FAST;
	LL i,N,a;
	cin>>N;
	map<LL,LL> m;
	VL v;
	rep(i,N){
		cin>>a;
		if(!m.count(a)){
			if(v.size()==3){
				cout<<"No"<<endl;
				return 0;
			}
			v.PB(a);
		}
		++m[a];
	}
	SORT(v);
	if(v[0]==0){
		if(v.size()==1){
			cout<<"Yes"<<endl;
			return 0;
		}
		if(v.size()==2)if(m[0]*2==m[v[1]]){
			cout<<"Yes"<<endl;
			return 0;
		}
	}else{
		if(v.size()==3&&(v[0]^v[1])==v[2]&&m[v[0]]==m[v[1]]&&m[v[2]]==m[v[1]]){
			cout<<"Yes"<<endl;
			return 0;
		}
	}
	cout<<"No"<<endl;
}