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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

//conversion
//------------------------------------------
inline int toInt(string s) {int v; istringstream sin(s);sin>>v;return v;}
template<class T> inline string toString(T x) {ostringstream sout;sout<<x;return sout.str();}

//math
//-------------------------------------------
template<class T> inline T sqr(T x) {return x*x;}

//typedef
//------------------------------------------
typedef long long LL;
typedef pair<int, int> PII;
typedef pair<LL,LL> PLL;
typedef map<int,int> MII;
typedef queue<int> QI;
typedef queue<PII> QPII;
typedef stack<int> SI;
typedef stack<PII> SPII;
typedef deque<int> DI;
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<LL> VLL;
typedef vector<VLL> VVLL;
typedef vector<bool> VB;
typedef vector<VB> VVB;
typedef vector<double> VD;
typedef vector<VD> VVD;
typedef vector<string> VS;
typedef vector<VS> VVS;
typedef vector<char> VC;
typedef vector<VC> VVC;
typedef vector<PII> VPII;
typedef vector<PLL> VPLL;
typedef priority_queue<int> PQGI; //大きい順
typedef priority_queue<int, VI, greater<int>> PQLI;
typedef priority_queue<PII> PQGP; 
typedef priority_queue<PII, VPII, greater<int>> PQLP;

//container util
//------------------------------------------
#define ALL(a)  (a).begin(),(a).end()
#define RALL(a) (a).rbegin(),(a).rend()
#define PB push_back
#define PF push_front
#define POB pop_back
#define POF pop_front
#define MP make_pair
#define SZ(a) int((a).size())
#define SQ(a) ((a)*(a))
#define EACH(i,c) for(typeof((c).begin()) i=(c).begin(); i!=(c).end(); ++i)
#define EXIST(s,e) ((s).find(e)!=(s).end())
#define SORT(c) sort((c).begin(),(c).end())
#define SORTR(c) sort((c).rbegin(), (c).rend())
#define LB lower_bound
#define UB upper_bound
#define NEXP next_permutation
#define FI first
#define SE second

//repetition
//------------------------------------------
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n)  FOR(i,0,n)
#define FORR(i,a,b) for(int i = (b-1);i>=a;i--)
#define REPR(i,n) FORR(i,0,n)
#define BFOR(bit,a,b) for(int bit = (a); bit < (1<<(b)); ++bit)
#define BREP(bit,n) BFOR(bit,0,n)

//constant
//--------------------------------------------
const double EPS = 1e-10;
const double PI  = acos(-1.0);
const int INF = INT_MAX/2;
const LL LINF = LLONG_MAX/3;
const int RINF = INT_MIN/2;
const LL RLINF = LLONG_MIN/3;
const LL MOD = 1e9+7;
const LL MODD = 998244353;
const int MAX = 510000;

inline bool Eq(double a, double b) { return fabs(b - a) < EPS; }

//other
//-------------------------------------------
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }

#define COUT(x) cout << (x) << endl
#define COUT2(x,y) cout << (x) << " " << (y) << endl
#define PR(x) cout << (x)
#define ENDL cout << endl
#define SPACE cout << " "
#define BC(x) __builtin_popcountll(x)

void cYes(){ COUT("Yes");exit(0); }
void cNo(){ COUT("No");exit(0); }
void cYES(){ COUT("YES");exit(0); }
void cNO(){ COUT("NO");exit(0); }
void cyes(){ COUT("yes");exit(0); }
void cno(){ COUT("no");exit(0); }
void cMINU(){ COUT(-1);exit(0); }
void cZERO(){ COUT(0);exit(0); }

VI dx = {1,0,-1,0,1,1,-1,-1};
VI dy = {0,1,0,-1,1,-1,1,-1};
VC dc = {'R','D','L','U'};
VS ds = {"R","D","L","U","RD","RU","LD","LU"};

LL Gcd(LL a, LL b) {return __gcd(a,b);}
LL Lcm(LL a, LL b) {return a/Gcd(a,b)*b;}

LL ModPow(LL A, LL N, LL M){
  LL res = 1;
  while(N>0){
    if(N&1) res = res * A % M;
    A = A * A % M;
    N >>= 1;
  }
  return res;
}

LL fac[MAX], finv[MAX], inv[MAX];

void ConInit(LL M) {
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < MAX; i++){
        fac[i] = fac[i - 1] * i % M;
        inv[i] = M - inv[M%i] * (M / i) % M;
        finv[i] = finv[i - 1] * inv[i] % M;
    }
}

LL COM(int n, int k, LL M){
    if (n < k) return 0;
    if (n < 0 || k < 0) return 0;
    return fac[n] * (finv[k] * finv[n - k] % M) % M;
}

template<class T> void VCout(vector<T> &v){
  int N = SZ(v);
  REP(i,N){
    cout << v.at(i);
    if(i == N-1) cout << endl;
    else cout << " ";
  }
  return;
}




int main() {
  
  cin.tie(0);
  ios::sync_with_stdio(false);
  cout << fixed << setprecision(12);
  
  
  int N;
  cin >> N;
  MII mp;
  REP(i,N){
    int a;
    cin >> a;
    mp[a]++;
  }
  
  bool ans = true;
  
  if(N%3 == 0){
    int k = N/3;
    if(SZ(mp) > 3) ans = false;
    
    if(SZ(mp) == 3){
      int del = 0;
      
      for(auto p : mp){
        del ^= p.FI;
        if(p.SE != k) ans = false;
      }
      if(del != 0) ans = false;
    }
    
    if(SZ(mp) == 2){
      if(!mp.count(0)) ans = false;
      else if(mp[0] != k) ans = false;
    }
  }else if(SZ(mp) == 1){
    if(!mp.count(0)) ans = false;
  }else{
    ans = false;
  }
  
  if(ans) cYes();
  else cNo();
  
  
  return 0;
}