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 JavaScript (node.js v5.12) (judge verdict: Accepted)

var obj = init();
Main();
function nextInt(){return myconv(next(),1);}
function nextStrArray(){return myconv(next(),2);}//半角スペース分割
function nextIntArray(){return myconv(next(),4);}//半角スペース + 数値化
function nextCharArray(){return myconv(next(),6);}//1文字分割
function next(){return obj.next();}
function hasNext(){return obj.hasNext();}
function myin(){return require("fs").readFileSync("/dev/stdin", "utf8").trim().split("\n");}
function init(){
  var inputFile = myin();
  var returnObj = {
    "list" : inputFile, "index" : 0, "max" : inputFile.length,
    "hasNext" : function(){return (this.index < this.max);},
    "next" : function(){if(!this.hasNext()){return "";}else{var returnInput = this.list[this.index];this.index++;return returnInput;}}
  };
  return returnObj;
}
function myout(t){console.log(t);}//標準出力
function myerr(t){console.error(t);}//標準エラー出力
//[no]要素の扱い。数値型
//不明値、異常時:引数そのまま返す　1:数値へ変換
//2:半角SPで分割 4:半角SPで分割し、数値配列へ
//6:1文字で分割  7:1文字で分割し、数値配列へ
//8:半角SPで結合 9:改行で結合 0:文字なしで結合
function myconv(i,no){try{switch(no){case 1:return parseInt(i);case 2:return i.split(" ");case 4:return i.split(" ").map((a)=>Number(a));case 6:return i.split("");case 7:return i.split("").map((a)=>Number(a));case 8:return i.join(" ");case 9:return i.join("\n");case 0:return i.join("");default:return i;}}catch(e){return i;}}
function Main(){
  var N = nextInt();
  var list = nextIntArray();
  var count = {};
  for(var i = 0; i < N; i++){
    if(count[list[i]] == null){
      count[list[i]] = 1;
    }else{
      count[list[i]]++;
    }
  }
  var keys = Object.keys(count);
  myerr(keys);
  if(keys.length > 3){
    myout("No");
  }else if(keys.length == 3){
    
    if((keys[0] ^ keys[1]) == keys[2] && count[keys[0]] == count[keys[1]] && count[keys[1]] == count[keys[2]]){
      myout("Yes");
    }else{
      myout("No");
    }
  }else if(keys.length == 2){
    if(keys.indexOf("0") != -1){
      
      var zeroCount = count["0"];
      var notZeroCount;
      if(keys[0] != 0){
         notZeroCount = count[keys[0]];
      }else{
         notZeroCount = count[keys[1]];
      }
      myerr(zeroCount);
      myerr(notZeroCount);
      if(notZeroCount / 2 == zeroCount){
        myout("Yes");
      }else{
        myout("No");
      }
    }else{
      myout("No");
    }
  }else{
    if(keys[0] == 0){
      myout("Yes");
    }else{
      myout("No");
    }
  }
}