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**Problem :-**Construct an N input XOR Gate. An XOR Gate returns 1 if odd number of its inputs are 1, otherwise 0.

**Input:**

The first line of input takes the number of test cases, T. Then T test cases follow.Each test case consists of 2 lines. The first line of each test case takes the number of inputs to the XOR Gate, N. The second line of each test case takes N space separated integers denoting the inputs to the XOR Gate. Note that the inputs can be either 1's or 0's.

**Output:**

For each test case on a new line print the output of the N input XOR Gate.

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**Solution :-**

**#include <bits/stdc++.h>**

using namespace std;

int findOdd(int arr[], int n)

{

int res = 0, i;

for (i = 0; i < n; i++)

res ^= arr[i];

return res;

}

int main()

{

int t;

cin>>t;

while(t--)

{

int n,i,j,sum,res=0;

cin>>n;

int arr[n];

for(i=0;i<n;i++)

cin>>arr[i];

printf ("%d\n", findOdd(arr, n));

}

return 0;

}

**Output :-**

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