# Problem Link:

Practice

* Author:* Sakchi Agarwal

*Sparsh Kedia*

**Author:***Ayush Shukla*

**Editorialist:**# DIFFICULTY:

Cakewalk

# PREREQUISITES:

Basic Mathematics

# PROBLEM:

You are given an array of N elements. Each element of the array is prime. Find the number of pairs that can be formed such that difference of the elements of the pair is odd.

# QUICK EXPLANATION:

The one element of pair must be even. And we have only one even prime number, i.e. , 2.

# EXPLANATION:

We know that total number of element in array is N. Lets keep count of number 2 in array, denoted as, count.

Now the total number of pair such that their difference is an odd number is (count)*(N-count).

## Setter's Solution

```
#include <bits/stdc++.h>
using namespace std;
#define ll long long
int main()
{ ll t;
cin >> t;
while (t--)
{
ll n;
cin>>n;
ll a[n];
ll count=0;
for(ll i=0;i<n;i++)
{
cin>>a[i];
if(a[i]%2==0)count++; //a[i] is 2
}
cout<<count*(n-count)<<endl;
}
return 0;
}
```

## Tester's Solution

```
import math
t = int(input())
while t>0 :
n = int(input())
cnt2 = 0
l = input().split()
for i in range(0,n):
if(l[i]=='2'):
cnt2+=1
print((n-cnt2)*cnt2)
t = t-1
```

Feel free to share your approach. Happy Coding