find the odd occurence

/*
Given an array of positive integers where all numbers occur even number of times except one number which occurs odd number of times. Find the 
number.

Input:
The first line of input contains a single integer T denoting the number of test cases. Then T test cases follow. Each test case consist of two
lines. The first line of each test case consists of an integer N, where N is the size of array. The second line of each test case contains N 
space separated integers denoting array elements.

 

Output:
Corresponding to each test case, print the number which occur odd number of times. If no such element exists, print 0.

Constraints:
1 ≤ T ≤ 100
1 ≤ N ≤ 107
1 ≤ A[i] ≤ 106

Example:
Input:
1
5
8 4 4 8 23
Output:
23

Explanation:
Testcase 1: 23 is the element which occurs odd number of times.
*/

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

int main() {
	int t; cin>>t; while(t--){
	    int n; cin>>n;
	    int a[n],flag=0;
	    for(int i=0; i<n; i++){
	        cin>>a[i];
	    }
	    flag=a[0];
	    for(int i=1; i<n; i++){
	        flag=flag^a[i];
	    }
	    cout<<flag<<endl;
	}
	return 0;
}