Power M divisibility by 7: PMD7(Assignment)

link: https://www.codechef.com/CU1AP0007/problems/PMD7

You are given two integers: X and M. Now, you need to replace each digit of X (let’s call them di) with dMimod10. Let’s call the new number Y and reverse it (explained in note 3). Check whether Y is divisible by 7.

Note 1: dMi=di⋅di⋅…⋅di (M times) – definition of di to the power of M. Special case: dMi=1, when M=0
Note 2: Xmod10 is the remainder of division X by 10, where X is some integer number.

Note 3: Reversing an integer is an operation that reverse the order of its digits and erase leading zeros. For example 123 when reversed becomes 321 and 450 when reversed becomes 54.

Input Format
The first line of the input contains a single integer: T – representing the number of test cases
Each of the test cases contains a line with two space-separated integers: X and M
Output Format
For each test case output a line containing either “YES” or “NO” depending on whether Y is divisible by 7.

You may print each character of the string in uppercase or lowercase (for example: the strings “yes”, “YeS”, “YES” will be treated as the same strings).

Constraints
1≤T≤500
1≤X≤109
0≤M≤100
Sample Input 1
1
123 2
Sample Output 1
NO
Explanation
After squaring all the digits the number we get is 149 and when we reverse it, it becomes 941. 941 is not divisible by 7.

Sample Input 2
1
123 4
Sample Output 2
YES

Solution:

#include <bits/stdc++.h>
#define ll long long int
using namespace std;

ll powr(ll x, ll n)
{
    ll res = 1;
    while (n)
    {
        if (n & 1)
        res = res * x % 10;
        n = n / 2;
        x = x * x % 10;
    }
    return res;
}
int main()
{
    int T;
    cin >> T;
    while (T--)
    {
        ll x, m;
        cin >> x >> m;
        ll ld, sum = 0;
        ll z;
        while (x > 0)
        {
            ld = x % 10;
            z = powr(ld, m);
            sum = 10 * sum + z;
            x /= 10;
        }
        if (sum % 7 == 0)
        cout << "YES" << endl;
        else
        cout << "NO" << endl;
    }
}