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.lang.reflect.Array;
  import java.util.*;
  
  public class MainClass {
      static long mod = (long)1e9 + 7;
      static long[] powers = new long[200000 + 10];
      public static void main(String[] args) {
          powers[0] = 1L;
          for (int i = 1; i < powers.length; i++) {
              powers[i] = (powers[i - 1] * 2L) % mod;
          }
          Reader in = new Reader(System.in);
          StringBuilder stringBuilder = new StringBuilder();
          int t = in.nextInt();
          while (t-- > 0) {
              int n = in.nextInt(), k = in.nextInt();
              if (k == 0) {
                  stringBuilder.append("1\n");
                  continue;
              }
              long[][] dp = new long[k][2];
              dp[k - 1][0] = (powers[n - 1] - (n % 2 == 0? 1L:0) + (n % 2 == 1? 1L:0) + mod) % mod;
              dp[k - 1][1] = (n % 2 == 0? 1: 0);
              for (int i = k - 2; i >= 0; i--) {
                  dp[i][0] = ((powers[n - 1] - (n % 2 == 0? 1L:0) + (n % 2 == 1? 1L:0) + mod) % mod * dp[i + 1][0]) % mod;
                  dp[i][1] = (((n % 2 == 0? 1L: 0) * binPow(powers[n], 1L * (k - 1 - i))) % mod + (((powers[n - 1] - (n % 2 == 0? 1L:0) + (n % 2 == 1? 1L:0) + mod) % mod) * dp[i + 1][1]) % mod) % mod;
              }
              long ans = (dp[0][0] + dp[0][1]) % mod;
              stringBuilder.append(ans).append("\n");
          }
          System.out.println(stringBuilder);
      }
      public static long binPow(long a, long n) {
          if (n == 0) {
              return 1L;
          }
          long res = binPow(a, n / 2);
          res = (res * res) % mod;
          if (n % 2 == 1L) {
              res = (res * a) % mod;
          }
          return res;
      }
  }
  class Reader {
      public BufferedReader reader;
      public StringTokenizer tokenizer;
  
      public Reader(InputStream stream) {
          reader = new BufferedReader(new InputStreamReader(stream), 32768);
          tokenizer = null;
      }
  
      public String next() {
          while (tokenizer == null || !tokenizer.hasMoreTokens()) {
              try {
                  tokenizer = new StringTokenizer(reader.readLine());
              } catch (IOException e) {
                  throw new RuntimeException(e);
              }
          }
          return tokenizer.nextToken();
      }
  
      public int nextInt() {
          return Integer.parseInt(next());
      }
  
      public long nextLong() {
          return Long.parseLong(next());
      }
  
  }