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: #include <bits/stdc++.h>
  using namespace std;
  #pragma GCC optimize("O3")
  const long long int mod = 1e9 + 7;
  long long int powmod(long long int x, long long int y) {
    long long int t;
    for (t = 1; y; y >>= 1, x = x * x % mod)
      if (y & 1) t = t * x % mod;
    return t;
  }
  long long int gcd(long long int x, long long int y) {
    return y ? gcd(y, x % y) : x;
  }
  long long int lcm(long long int x, long long int y) {
    return x * (y / gcd(x, y));
  }
  long long int modd(long long int a) { return (a % mod + mod) % mod; }
  double findMod(double a, double b) {
    double mods;
    if (a < 0)
      mods = -a;
    else
      mods = a;
    if (b < 0) b = -b;
    while (mods >= b) mods = mods - b;
    if (a < 0) return -mods;
    return mods;
  }
  vector<long long int> sieve(int n) {
    int* arr = new int[n + 1]();
    vector<long long int> vect;
    for (long long int i = 2; i <= n; i++)
      if (arr[i] == 0) {
        vect.push_back(i);
        for (long long int j = 2 * i; j <= n; j += i) arr[j] = 1;
      }
    return vect;
  }
  long long int add(long long int a, long long int b) {
    return modd(modd(a) + modd(b));
  }
  long long int mul(long long int a, long long int b) {
    return modd(modd(a) * modd(b));
  }
  int smask(int i, int pos) { return (i | (1 << pos)); }
  int clmask(int i, int pos) { return (i & (~(1 << pos))); }
  bool chmask(int i, int pos) { return (i & (1 << pos)) != 0; }
  double cordist(pair<double, double> a, pair<double, double> b) {
    return sqrt(((a.first - b.first) * (a.first - b.first)) +
                ((a.second - b.second) * (a.second - b.second)));
  }
  long long binpow(long long a, long long b) {
    if (b <= 0) return 1;
    long long res = binpow(a, b / 2);
    if (b % 2)
      return mul(mul(res, res), a);
    else
      return mul(res, res);
  }
  long long int divs(long long int a, long long int b) {
    return a * binpow(b, mod - 2);
  }
  long long int n, m, q, r, i, j;
  vector<long long int> arr;
  void solve() {
    cin >> n >> q;
    if (n & 1) {
      cout << binpow(binpow(2, n - 1) + 1, q) % mod << '\n';
    } else {
      long long int sparse = binpow(binpow(2, n - 1) - 1, q);
      long long int ans = (binpow(2, (long long int)n * q) - sparse);
      ans *= binpow((binpow(2, n - 1) + 1), mod - 2);
      ans += sparse;
      ans %= mod;
      while (ans < 0) ans += mod;
      cout << ans << '\n';
    }
  }
  signed main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    int k = 1;
    cin >> k;
    q = 0;
    while (k--) {
      q++;
      solve();
    }
    return 0;
  }
  