## Problem

Gru has not been in the limelight for a long time and is, therefore, planning something particularly nefarious. Frustrated by his minions’ incapability which has kept him away from the limelight, he has built a transmogrifier — a machine which mutates minions.

Each minion has an intrinsic characteristic value (similar to our DNA), which is an integer. The transmogrifier adds an integer **K** to each of the minions’ characteristic value.

Gru knows that if the new characteristic value of a minion is divisible by 7, then it will have Wolverine-like mutations.

Given the initial characteristic integers of **N** minions, all of which are then transmogrified, find out how many of them become Wolverine-like.

### Input Format:

The first line contains one integer, **T**, which is the number of test cases. Each test case is then described in two lines.

The first line contains two integers **N** and **K**, as described in the statement.

The next line contains **N** integers, which denote the initial characteristic values for the minions.

### Output Format:

For each testcase, output one integer in a new line, which is the number of Wolverine-like minions after the transmogrification.

### Constraints:

**1 ≤ T ≤ 100****1 ≤ N ≤ 100****1 ≤ K ≤ 100**- All initial characteristic values lie between 1 and 10
^{5}, both inclusive.

### Example

Input:1 5 10 2 4 1 35 1Output:1

### Explanation:

After transmogrification, the characteristic values become {12,14,11,45,11}, out of which only 14 is divisible by 7. So only the second minion becomes Wolverine-like.

## Mutated Minions- CodeChef Solution in CPP

#include <iostream> using namespace std; void solve_test() { int n, k, current; cin >> n >> k; int arr[n]; int i; for (i = 0; i < n; i++) cin >> arr[i]; int result = 0; for (i = 0; i < n; i++) { current = arr[i] + k; if (current % 7 == 0) result++; } cout << result << endl; } int main() { int t; cin >> t; while (t--) solve_test(); return 0; }

** Disclaimer: The above Problem ( Mutated Minions ) is generated by** CodeChef but the solution is provided by

**Chase2learn**.This tutorial is only for

**Educational**and

**Learning**purpose.