The XOR Gate

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.

Constraints:

1<=T<=100
1<=N<=100

Example:

Input:

3
2
1 1
3
1 0 1
4
1 1 1 0

Output:
0
0
1

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SIMPLE c++ IMPLEMENTATION :
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#include<iostream>
using namespace std;
int result(int [],int ) ;
int main()
 {
int t,arr[100],i,no ;
cin>>t ;
while(t--)
{
   cin>>no ;
   for(i=0;i<no;i++)
       cin>>arr[i] ;
     
   cout<<result(arr,no) ;
   cout<<endl ;
}
return 0;
}

int result(int arr[],int no)
{
    int i ,ans ;
    ans=0 ;
   
    for(i=0;i<no;i++)
        ans=ans^arr[i] ;    // ^ (bitwise XOR) Takes two numbers as operand and does XOR on every bit                                    // of two numbers. The result of XOR is 1 if the two bits are different.
   
    return ans ;
}

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