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.io.*;
  import java.util.*;
  
  public class Program{
      static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
      static PrintWriter pw = new PrintWriter(System.out);
      static StringTokenizer st = new StringTokenizer("");
      static long MOD = 1000000007;
      static long fact[];
      static int T, N, K;
      static long dp[][];
      static long inv(long x){
          long ret = 1;
          long e = MOD-2;
          while(e > 0){
              if((e&1) > 0) ret = ret * x % MOD;
              x = x * x % MOD;
              e >>= 1;
          }
          return ret;
      }
      static long nCr(int n, int r){
          return fact[n] * inv(fact[r] * fact[n-r] % MOD) % MOD;
      }
      public static void main(String args[]) throws IOException{
          fact = new long[250000];
          fact[0] = 1;
          for(int i = 1; i < 250000; i++){
              fact[i] = fact[i-1] * i % MOD;
          }
          T = nexti();
          while(T-- > 0){
              N = nexti();
              K = nexti();
              long val = 0;
              for(int i = 0; i < N; i += 2){
                  val = (val + nCr(N, i)) % MOD;
              }
              long p2n = 1;
              for(int i = 0; i < N; i++){
                  p2n = p2n * 2 % MOD;
              }
              dp = new long[K+1][2];
              for(int i = 0; i <= K; i++) for(int j = 0; j < 2; j++) dp[i][j] = 0;
              dp[0][0] = 1;
              for(int i = 1; i <= K; i++){
                  dp[i][1] = dp[i-1][1] * p2n % MOD;
                  dp[i][0] = val * dp[i-1][0] % MOD;
                  if(N % 2 == 0){
                      dp[i][1] = (dp[i][1] + dp[i-1][0]) % MOD;
                  }else{
                      dp[i][0] = (dp[i][0] + dp[i-1][0]) % MOD;
                  }
              }
              pw.println((dp[K][0] + dp[K][1]) % MOD);
          }
          br.close(); pw.close();
      }
      static String next() throws IOException{
          while(!st.hasMoreTokens()) st = new StringTokenizer(br.readLine());
          return st.nextToken();
      }
      static int nexti() throws IOException{
          return Integer.parseInt(next());
      }
  }