Power of large Numbers

//  I can't tell you what it really is,
// I can only tell you what it feels like.
/*
#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;
typedef tree<int,null_type,less<int>,rb_tree_tag,
tree_order_statistics_node_update> indexed_set;
*/
#include <bits/stdc++.h>
using namespace std;
#define fast() ios_base::sync_with_stdio(0);cin.tie(NULL);cout.tie(NULL)
typedef long long  ll;
typedef unsigned long long ull;
typedef long double ld;
const long long int N=1e6 + 1;
const long int x = 1e7;
#define pb(x)  push_back(x)
#define sd(x) scanf("%d", &(x))
#define I(x) scanf("%lld" , &x)
#define A(s) s.begin(),s.end()
#define L(x) (int)((x).size())
#define V vector<int>
#define VL vector<ll>
#define S set<int>
#define SL set<ll>
#define Ml multiset < int >
#define ML multiset < ll >
#define in    freopen ("in.txt", "r", stdin);
#define out   freopen ("out.txt", "w", stdout);
#define f(i,p,n) for(int i=(p);i<n;i++)
#define M     map < ll , ll >
#define F     first
#define SS    second
const long int mod = 1e9 + 7 ;
ll po(ll a,ll b,ll mod) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
//https://tp-iiita.quora.com/
/*
Fermat's Little Theorem: If P is a prime number, then for every integer A, according to Fermat’s little theorem,

AP≡A(modP)

Further, if A is not divisible by P,

Ap−1≡1(modP)

Solution: We cannot use modular exponentiation directly in this problem because BC is itself a very large number and will not fit even in long long. We’ll use Fermat’s little theorem and reduce the problem.

Let	X=BC,

Now we have to calculate AX.
Let	X=BC,

Now we have to calculate AX.

Suppose	Q=X/(M−1)	and	R=X%(M−1)

So,	AX=AQ∗(M−1)+R

              =AQ∗(M−1)∗AR

              =(AM−1)Q∗AR

              =1Q∗AR

[Using Fermat’s little theorem, M is prime and A is not divisible by M, so AM−1≡1(modM)]

=>AX=AR

Hence, for computing ABC, we first have to compute R=X%(M−1)=BC%(M−1) using modular exponentiation and after that we have to compute AR%M by using modular exponentiation again. Time Complexity will be O(logC+logM). */
int main(int argc, char const *argv[]){
   fast();
   #ifndef ONLINE_JUDGE
   //freopen("in.txt", "r", stdin);
   //freopen("out.txt", "w", stdout);
   #endif
  	int t;
  	cin >> t;
  	while(t--){
  		string a , b;
  		cin >> a >> b;
  		ull res = 0 , num = 0;
  		f(i,0,L(a)){
  			res = (res * 10 + (a[i] - '0'))%mod;
  		}
  		f(i,0,L(b)){
  			num = (num * 10 + (b[i] - '0'))% (mod - 1);
  		}
  		ull temp;
  		temp = po(res , num , mod);
        cout << temp << '\n';
  	}





  	return 0;
}