For this Question: Given any number S ending with 1. What will be the last 2 digits of the resultant number when raised to the power P?The first line of input contains a single integer T denoting the number of test cases. Then T test cases follow. Each test case consists of string S and power P.

Constraints:
1 <= T <= 2*10^3
1 <= |S| <= 60
1 <= P <= 10^6Print the last 2 digit of the resultant. If answer is 1 digit number X, print 0X.Input:
2
1231 213
43591 194

Output:
91
61

Explanation:
Testcase 1: When (1231)^213 is calculated the last 2 digits appeared to be 91., I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;

#define pu push_back
#define fi first
#define se second
#define mp make_pair
#define int long long
#define pii pair<int,int>
#define mm (s+e)/2
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(int i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define sz 200000
signed main()
{
	
		int t;
		cin>>t;
		while(t>0)
		{
			t--;
			string s;
			cin>>s;
			int yy;
			cin>>yy;
			int n=s.size();
			if(n==1) {
				cout<<"01"<<endl;
				continue;
			}
			int p=s[n-2]-'0';
			int q=(p*yy)%10LL;
			//if(q==0 ) cout<<1<<endl;
			 cout<<q<<1<<endl;



		}
	
}, In this Programming Language: C++, Now tell me if this Code is compilable or not?