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.*;
  
  public class _1557_C {
      static final int MOD = 1000000007;
      static final long[] fact = new long[200005];
      static final long[] inv = new long[200005];
      static final long[] pow = new long[200005];
      public static void main(String[] args) throws IOException {
          fact[0] = 1;
          pow[0] = 1;
          for(int i = 1; i < 200005; i++) {
              fact[i] = modmult(fact[i - 1], i);
              inv[i] = modinv(fact[i]);
              pow[i] = modmult(pow[i - 1], 2);
          }
          BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
          PrintWriter out = new PrintWriter(System.out);
          int t = Integer.parseInt(in.readLine());
          while(t-- > 0) {
              StringTokenizer line = new StringTokenizer(in.readLine());
              int n = Integer.parseInt(line.nextToken());
              int k = Integer.parseInt(line.nextToken());
              //binom[i] stores n choose i
              long[] binom = new long[n + 1];
              binom[0] = 1;
              long num = 1;
              long sum = 0;
              for(int i = 0; i < n; i++) {
                  num = modmult(num, n - i);
                  binom[i + 1] = modmult(num, inv[i + 1]);
                  if(i % 2 == 0) {
                      sum = modadd(sum, binom[i]);
                  }
              }
              //dp[i] stores number of winning arrays with k = i
              long[] dp = new long[k + 1];
              dp[0] = 1;
              for(int i = 1; i <= k; i++) {
                  if(n % 2 == 0) {
                      dp[i] = modadd(dp[i], binpow(pow[i - 1], n));
                  }else {
                      dp[i] = modadd(dp[i], dp[i - 1]);
                  }
                  dp[i] = modadd(dp[i], modmult(sum, dp[i - 1]));
              }
              out.println(dp[k]);
          }
          in.close();
          out.close();
      } 
      static long modadd(long a, long b) {
          return (a + b + MOD) % MOD;
      }
      static long modmult(long a, long b) {
          return a * b % MOD;
      }
      static long modinv(long a) {
          return binpow(a, MOD - 2);
      }
      static long binpow(long a, long b) {
          if(b == 0) {
              return 1;
          }
          long small = binpow(a, b / 2);
          if(b % 2 == 0) {
              return modmult(small, small);
          }else {
              return modmult(modmult(small, small), a);
          }
      }
  }
  