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)

#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,fma,abm,mmx,avx,tune=native")
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp> // Common file  
#include <ext/pb_ds/tree_policy.hpp>  
#include <functional> // for less  
#include <iostream>  
using namespace __gnu_pbds;  
using namespace std;
#define ll long long
typedef tree<pair<ll,ll>, null_type, less<pair<ll,ll>>, rb_tree_tag,tree_order_statistics_node_update>  ordered_set;
using namespace std;
 
#define fr(i,a,b) for (int i = (a), _b = (b); i <= _b; i++)
#define frr(i,a,b) for (int i = (a), _b = (b); i >= _b; i--)
#define rep(i,n) for (long long i = 0, _n = (n); i < _n; i++)
#define repr(i,n) for (long long i = (n) - 1; i >= 0; i--)
#define foreach(it,ar) for ( typeof(ar.begin()) it = ar.begin(); it != ar.end(); it++ )
#define fill(ar,val) memset(ar, val, sizeof(ar))
#define fill0(ar) fill((ar), 0)
#define fillinf(ar, n) fr(i,0,(n)) ar[i] = INF
#define debug(x) cout<<#x<<": "<<x<<endl
#define arr1d(a,n) cout << #a << " : "; fr(_,1,n) cout << a[_] << ' '; cout << endl;
#define arr1d0(a,n) cout << #a << " : "; rep(_,n) cout << a[_] << ' '; cout << endl;
#define arr2d(a,n,m) cout << #a << " :" << endl; fr(_,1,n) {fr(__,1,m) cout << a[_][__] << ' '; cout << endl;}
#define arr2d0(a,n,m) cout << #a << " :" << endl; rep(_,n) {rep(__,m) cout << a[_][__] << ' '; cout << endl;}
/*Author Ritick Goenka || ritick(codechef) ||ritick(codeforces) */
/*IIT Roorkee = <3 */
#define ull unsigned long long
#define ld long double
#define ui unsigned int
#define all(ar) ar.begin(), ar.end()
#define pb push_back
#define mp make_pair
#define ff first
#define ss second
#define y0 yyyyyy0
auto clk=clock();
#define y1 yyyyyy1
#define setb __builtin_popcount
#define BIT(n) (1<<(n))
#define SQR(x) ((x) * (x))
#define CUBE(x) ((x) * (x) * (x))
#define LSOne(S) (S) & (-S)
inline bool EQ(double a, double b) {return fabs(a - b) < 1e-9;}
 
typedef pair<int, int> ii;
typedef pair<int, ii> iii;
typedef vector<ii> vii;
typedef vector<int> vi;
typedef vector<string> vs;
 
template<typename T>inline T gcd(T a, T b){if (b == 0)return a;else return gcd(b, a % b);}
// template<typename T>inline T lcm(T a, T b){return (a * b) / gcd(a, b);}
template<typename T> string toStr(T x) {stringstream st; st << x; string s; st >> s; return s;}
template<class T>
void splitStr(const string &s, vector<T> &out)
{
    istringstream in(s);
    out.clear();
    copy(istream_iterator<T>(in), istream_iterator<T>(), back_inserter(out));
}
 
inline int two(int n) {return 1 << n;}
inline int isOnBit(int n, int b) {return (n >> b) & 1;}
inline void onBit(int & n, int b) {n |= two(b);}
inline void offBit(int & n, int b) {n &= ~two(b);}
inline int lastBit(int n) {return n & (-n);}
inline int cntBit(int n) {int res = 0; while(n && ++res) n -= n &(-n); return res;}
 
const int dx4[] = {-1, 0, 1, 0};
const int dy4[] = {0, 1, 0, -1};
const int dx8[] = {-1, 0, 1, 0, -1, -1, 1, 1};
const int dy8[] = {0, 1, 0, -1, -1, 1, -1, 1};
 
#define INP "test.inp"
#define OUT "test.out"
 
#define PI 3.1415926535897932385
#define INF 10000000000000000ll
#define EPS 1e-7
#define MAXN 100000
#define MOD 1000000007
#define dec decr
#define endl '\n'
 
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
ll gcd(ll a, ll b) 
{ 
    if (a == 0) 
        return b; 
    return gcd(b % a, a); 
} 
ll modexp(ll x, ll y, ll p)
{
    ll res = 1;   
    x = x % p;  
    while (y > 0)
    {
        if (y %2==1)    
            res = (res*x) % p;
        y = y>>1;     
        x = (x*x) % p; 
    }
    return res;
}
int main () 
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout<<fixed;
    cout<<setprecision(9);
    #ifdef rg
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
    #endif
    ll n;
    cin>>n;
    vector<ll>v;
    set<ll>s;
    map<ll,ll>m;
    ll ze=0;
    rep(i,n)
    {
        ll x;
        cin>>x;
        v.pb(x);
        if(x!=0)
        s.insert(x);
        ze++;
        if(m.count(x)==0)
        {
            m[x]=1;
        }
        else
        {
            m[x]++;
        }
    }
    if(s.empty())
    {
        cout<<"Yes";
        return 0;
    }
    if(s.size()==1&&n==ze*3)
    {
        cout<<"Yes";
        return 0;
    }
    if(s.size()!=3)
    {
        cout<<"No";
        return 0;
    }
    ll z=0;
    ll xx=*s.begin();
    xx=m[xx];
    vector<ll>xz;
    for(auto x:s)
    {
        xz.pb(x);
        if(m[x]!=xx)
        {
            cout<<"No";
        return 0;
        }
    }
    xx=v[0];
    if(v[0]^v[2]==v[1]&&v[2]==v[0]^v[1]&&v[1]==v[2]^v[0])
    {
        cout<<"Yes";
        return 0;
    }
    cout<<"No";
        return 0;






    #ifdef rg
    cout<<endl<<endl<<"Time elapsed: "<<(double)(clock()-clk)/CLOCKS_PER_SEC<<endl;
    #endif
    return 0;
}