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.BufferedReader;
  import java.io.InputStreamReader;
  import java.io.PrintWriter;
  
  public class ProblemC {
  
      public BufferedReader scanner = new BufferedReader(new InputStreamReader(System.in));
      public PrintWriter out = new PrintWriter(System.out);
      public String[] in;
      int n, k;
      final int MAXN = (int)1e6;
      long[] pow2 = new long[MAXN];
  
      final long MOD = (long) (1e9+7);
  
      long fast_pow(long x, long y) {
          if (y == 0)
              return 1;
          if (y == 1)
              return x;
          long res =  fast_pow(x, y/2);
          res = res * res % MOD;
          if (y % 2 == 1) {
              return  res * x % MOD;
          }
          return res;
      }
  
      void solve(int n, int k) {
          long dp[] = new long[k+1];
          dp[0] = 1;
          for (int i = 1; i <= k; i++) {
              if (n % 2 == 0) {
                  dp[i] = (pow2[n-1]-1)*dp[i-1] + fast_pow(pow2[n], i-1);
              } else {
                  dp[i] = (pow2[n-1]+1)*dp[i-1];
              }
               dp[i] %= MOD;
          }
          out.println(dp[k]);
      }
  
      void process() throws Exception {
          int t = Integer.parseInt(scanner.readLine());
          pow2[0] = 1;
          for (int i = 1; i < MAXN; i++) {
              pow2[i] = (pow2[i-1] * 2) % MOD;
          }
          while (t-- != 0) {
              in = scanner.readLine().split(" ");
              n = Integer.parseInt(in[0]);
              k = Integer.parseInt(in[1]);
              solve(n, k);
          }
          out.flush();
      }
  
      public static void main(String[] args) throws Exception {
          ProblemC  app = new ProblemC();
          app.process();
      }
  }
  