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.Scanner;
  
  public class C {
      static long factorial[] = new long[(int) 2e5 + 3];
      static long inverseFactorial[] = new long[(int) 2e5 + 3];
      static int MOD = (int) 1e9 + 7;
  
      public static void main(String[] args) {
          factorial[0] = 1;
          inverseFactorial[0] = 1;
          for (int i = 1; i <= (int) 2e5; i++) {
              factorial[i] = mul(factorial[i - 1], i);
              inverseFactorial[i] = binpow(factorial[i], MOD - 2);
          }
          Scanner scn = new Scanner(System.in);
          int test = scn.nextInt();
          while (test-- > 0) {
              int n = scn.nextInt();
              int k = scn.nextInt();
              long[] dp = new long[k + 1];
              dp[k] = 1;
              long equal = 0;
              long moreThan = 0;
              if (n % 2 != 0)
                  equal = 1l;
              else moreThan = 1l;
              for (int i = 0; i < n; i += 2) {
                  equal = add(equal, nCr(n, i));
              }
  
              for (int i = k - 1; i >= 0; i--) {
                  dp[i] = mul(equal, dp[i + 1]);
                  dp[i] = add(dp[i], mul(moreThan, binpow(2, 1l*n * (k - i - 1))));
              }
              System.out.println(dp[0]);
          }
  
      }
  
      static long add(long a, long b) {
          if ((a += b) >= MOD)
              a -= MOD;
          else if (a < 0)
              a += MOD;
          return a;
      }
  
      static long mul(long x, long y) {
          return (1L * x * y) % MOD;
      }
  
      static long nCr(int n, int r) {
          if (r > n)
              return 0;
          return mul(mul(factorial[n], inverseFactorial[r]), inverseFactorial[n - r]);
      }
  
  
      public static long binpow(long x, long n) {
          assert (n >= 0);
          x %= MOD;  // note: MOD * MOD must be less than 2^63 to avoid ll overflow
          long res = 1;
          while (n > 0) {
              if (n % 2 == 1)  // if n is odd
                  res = res * x % MOD;
              x = x * x % MOD;
              n /= 2;  // divide by two
          }
          return res;
      }
  }
  