description: Moamen and Ezzat are playing a game. They create an array a of n non-negative integers where every element is less than 2^k.
  
  Moamen wins if a_1  \&  a_2  \&  a_3  \&  …  \&  a_n ≥ a_1 ⊕ a_2 ⊕ a_3 ⊕ … ⊕ a_n.
  
  Here \& denotes the [bitwise AND operation](https://en.wikipedia.org/wiki/Bitwise_operation#AND), and ⊕ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
  
  Please calculate the number of winning for Moamen arrays a.
  
  As the result may be very large, print the value modulo 1 000 000 007 (10^9 + 7).
  
  Input
  
  The first line contains a single integer t (1 ≤ t ≤ 5)— the number of test cases. 
  
  Each test case consists of one line containing two integers n and k (1 ≤ n≤ 2⋅ 10^5, 0 ≤ k ≤ 2⋅ 10^5).
  
  Output
  
  For each test case, print a single value — the number of different arrays that Moamen wins with.
  
  Print the result modulo 1 000 000 007 (10^9 + 7).
  
  Example
  
  Input
  
  
  3
  3 1
  2 1
  4 0
  
  
  Output
  
  
  5
  2
  1
  
  Note
  
  In the first example, n = 3, k = 1. As a result, all the possible arrays are [0,0,0], [0,0,1], [0,1,0], [1,0,0], [1,1,0], [0,1,1], [1,0,1], and [1,1,1].
  
  Moamen wins in only 5 of them: [0,0,0], [1,1,0], [0,1,1], [1,0,1], and [1,1,1].
solution: import java.util.*;
  import java.io.*;
  //import java.math.*;
  //https://www.hackerrank.com/goc-cdc-series-1
   public class Task{
  //  ..............code begins here..............
  
  static long mod=(long)1e9+7,mod1=998244353l,inf=(long)1e18+5;
  
  static void solve() throws  IOException{
      int[] z=int_arr();
      int n=z[0],k=z[1];
  
      if(n%2==1){
          long c=Modpow(2l,n-1,mod);
          c=Modpow((1l+c),k,mod);
          out.write(c+"\n");
      }
      else{
          long c=Modpow(2l,n-1,mod);
          c--;c+=mod;c%=mod;
          c=Modpow(c,k,mod);
          for(int i=1;i<=k;i++){
              int rk=k-i,left=i-1;
              long tmp=Modpow(2l,n-1,mod);
              tmp--;tmp+=mod;tmp%=mod;
              tmp=Modpow(tmp,left,mod);
              c+=tmp*Modpow(2l,(long)n*rk,mod)%mod;
              c%=mod;
          }
          out.write(c%mod+"\n");
      }
  }
  
  
  public static void main(String[] args) throws  IOException{
      assign();
      int t=int_v(read()),cn=1;
      while(t--!=0){
          //out.write("Case #"+cn+": ");
          solve();
          //cn++;
      }
      out.flush();
  }
  
  // taking inputs
  static BufferedReader s1;
  static BufferedWriter out;
  static String read() throws IOException{String line="";while(line.length()==0){line=s1.readLine();continue;}return line;}
  static int int_v (String s1){return Integer.parseInt(s1);}
  static long long_v(String s1){return Long.parseLong(s1);}
  static void sort(int[] a){List<Integer> l=new ArrayList<>();for(int x:a){l.add(x);}Collections.sort(l);for(int i=0;i<a.length;i++){a[i]=l.get(i);}}
  static int[] int_arr() throws IOException{String[] a=read().split(" ");int[] b=new int[a.length];for(int i=0;i<a.length;i++){b[i]=int_v(a[i]);}return b;}
  static long[] long_arr() throws IOException{String[] a=read().split(" ");long[] b=new long[a.length];for(int i=0;i<a.length;i++){b[i]=long_v(a[i]);}return b;}
  static void assign(){s1=new BufferedReader(new InputStreamReader(System.in));out=new BufferedWriter(new OutputStreamWriter(System.out));}
  //static String setpreciosion(double d,int k){BigDecimal d1 = new BigDecimal(Double.toString(d));return d1.setScale(k,RoundingMode.HALF_UP).toString();}//UP DOWN HALF_UP
  static int add(int a,int b){int z=a+b;if(z>=mod)z-=mod;return z;}
  static long gcd(long a,long b){if(b==0){return a;}return gcd(b,a%b);}
  static long Modpow(long a,long p,long m){long res=1;while(p>0){if((p&1)!=0){res=(res*a)%m;}p >>=1;a=(a*a)%m;}return res%m;}
  static long Modmul(long a,long b,long m){return ((a%m)*(b%m))%m;}
  static long ModInv(long a,long m){return Modpow(a,m-2,m);}
  //static long nck(int n,int r,long m){if(r>n){return 0l;}return Modmul(f[n],ModInv(Modmul(f[n-r],f[r],m),m),m);}
  //static long[] f;
  }