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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

/*
String[]の初期化、整数の型がlongになっているか
関数（splitとかcontainsとか）内では"+"とかは正規表現となされるので"\\+"とする
'+'とかは問題ない
配列のコピーはarr1=arr.clone()
HashSet<>[]はおのおの初期化した？？？？？
負の数の割り算は絶対値が小さいほうに切り捨てられるex.-1/2=0ので、ある値>kみたいな式だとバグらせやすいので注意
ある値>=kとして、切り上げた値をとるべき
toLowerCase()はアルファベット以外に対して行ってもエラーにはならない
ArrayDequeではpopとpushはStackと同じ操作になる
//import static java.lang.System.*; はしばらく使わなそう(使うとしてもexit(0)のためだけ)←return;でよさそう
最後にout.flush()またはout.close()をするのを忘れずに！
Point2D p2d=new Point2D.Double(d1,d2);
Line2D l2d=new Line2D.Double(x1,y1,x2,y2);
Line2D l2d_0=new Line2D.Double(x11,y11,x22,y22);
if(l2d.intersectsLine(l2d_0))
next_Permutationを使うときはソート済み配列かどうか確認！！
*/
//入力終了→Ctrl+D
import java.io.*;
import java.awt.*;
import java.lang.reflect.Array;
import java.util.*;
import java.awt.geom.*;
import static java.lang.Math.*;
public class Main {
    public static void main(String[] $) {
        int n=sc.nextInt();
        int[] a=new int[n];
        String[] bit=new String[n];
        int z=0;
        int lenmax=0;
        HashMap<Integer,Integer> map=new HashMap<>();
        for (int i = 0; i < n; i++) {
            a[i]=sc.nextInt();
            map.put(a[i],map.getOrDefault(a[i],0)+1);
            int t=a[i];
            int c=0;
            if(t==0)z++;
            bit[i]=Integer.toString(a[i],2);
            lenmax=max(bit[i].length(),lenmax);
        }
        int[][] bitarr=new int[n][lenmax];
        for (int i = 0; i < n; i++) {
            int t=a[i];
            int c=lenmax-1;
            while (t>0){
                bitarr[i][c]=t%2;
                t/=2;
                c--;
            }
        }
        String ans="No";
        if(z==n){
            ans="Yes";
        }else if(map.size()==3) {
            int[] key = new int[3];
            int counter = 0;
            int x = 0;
            for (Integer k : map.keySet()) {
                x ^= k;
                key[counter] = k;
                counter++;
            }
            if (x == 0 && map.get(key[0]) == map.get(key[1]) && map.get(key[1]) == map.get(key[2]) && map.get(key[2]) == map.get(key[0])) {
                ans = "Yes";
            }
        }else if(map.size()==2&&map.get(0)*2==n-map.get(0)) {
            ans="Yes";
        }
        out.println(ans);
        out.close();
    }

    static PrintWriter out = new PrintWriter(System.out);

    static class sc {
        static Scanner scanner = new Scanner(System.in);

        static String next() {
            return scanner.next();
        }

        static String nextLine() {
            return scanner.nextLine();
        }

        static int nextInt() {
            return Integer.parseInt(scanner.next());
        }

        static long nextLong() {
            return Long.parseLong(scanner.next());
        }

        static double nextDouble() {
            return Double.parseDouble(scanner.next());
        }

        static boolean hasNext() {
            return scanner.hasNext();
        }

        static boolean hasNextInt() {
            return scanner.hasNextInt();
        }

        static boolean hasNext(String str) {
            return scanner.hasNext(str);
        }
    }

}