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;
  const int Inf = 0x3f3f3f3f;
  const long long Lnf = 0x3f3f3f3f3f3f3f3f;
  template <typename T>
  inline T Min(T a, T b) {
    return a < b ? a : b;
  }
  template <typename T>
  inline T Max(T a, T b) {
    return a > b ? a : b;
  }
  template <typename T>
  inline T Abs(T x) {
    return x > 0 ? x : -x;
  }
  template <typename T>
  inline T GetMin(T tmp[], int l, int r) {
    T res = Inf;
    for (int i = l; i <= r; i++) res = min(res, tmp[i]);
    return res;
  }
  template <typename T>
  inline T GetMax(T tmp[], int l, int r) {
    T res = -Inf;
    for (int i = l; i <= r; i++) res = max(res, tmp[i]);
    return res;
  }
  template <typename T>
  inline T GetSum(T tmp[], int l, int r, int mod) {
    long long sum = 0;
    for (int i = l; i <= r; i++)
      if (mod == -1)
        sum += tmp[i];
      else
        sum = (sum + tmp) % mod;
    return sum;
  }
  template <typename T>
  inline void GetFrac(T tmp[], int l, int r, int mod) {
    tmp[l - 1] = 1;
    for (int i = l; i <= r; i++) tmp[i] = (tmp[i - 1] * i) % mod;
  }
  template <typename T>
  void Print(T tmp[], int l, int r, int type) {
    if (type == 1)
      for (int i = l; i <= r; i++) cout << tmp[i] << ' ';
    if (type == 2)
      for (int i = l; i <= r; i++) cout << tmp[i];
    if (type == 3)
      for (int i = l; i <= r; i++) cout << tmp[i] << '\n';
  }
  template <typename T>
  void GetPrime(T prime[], bool vis[], int up, int &top) {
    top = 0;
    for (int i = 2; i <= up; i++) {
      if (!vis[i]) prime[++top] = i;
      for (int j = 1; j <= top && i * prime[j] <= up; j++) {
        vis[i * prime[j]] = true;
        if (i % prime[j] == 0) break;
      }
    }
  }
  void checkf(bool flag, string Y, string N) {
    cout << ((flag == true) ? Y : N) << '\n';
  }
  long long qpow(long long x, long long p, long long mod) {
    long long base = x, res = 1 % mod;
    while (p != 0) {
      if (p & 1) res = (res * base) % mod;
      p >>= 1;
      base = (base * base) % mod;
    }
    return res;
  }
  const int Mod = 1e9 + 7;
  const int base = 131;
  const int maxn1 = 5e5 + 5;
  const int maxn2 = 5000 + 5;
  const int maxn3 = 500 + 5;
  const int len = 100 + 5;
  long long T, n, m, k;
  int a[maxn1];
  long long frac[maxn1];
  long long sum[maxn1];
  long long C(long long x, long long y) {
    if (y > x) return 0;
    long long ans = ((frac[x] * qpow(frac[y], Mod - 2, Mod)) % Mod *
                     qpow(frac[x - y], Mod - 2, Mod) % Mod);
    return ans;
  }
  long long Lucas(long long x, long long y) {
    if (y == 0)
      return 1;
    else
      return C(x, y) * Lucas(x / Mod, y / Mod) % Mod;
  }
  long long getfrac(long long l, long long r) {
    return frac[r] * qpow(frac[l - 1], Mod - 2, Mod) % Mod;
  }
  void solve() {
    scanf("%lld%lld", &n, &k);
    long long rest = 0, pre = 0, tot = 0;
    for (int i = 0; i < n; i += 2) pre = (pre + Lucas(n, i)) % Mod;
    if (n % 2 != 0) pre++;
    for (int i = 1; i < n; i += 2) tot = (tot + Lucas(n, i)) % Mod;
    for (int i = 1; i <= k; i++) {
      rest = (rest + (qpow(pre, i - 1, Mod) * tot % Mod *
                      qpow(2, n * (k - i), Mod) % Mod)) %
             Mod;
    }
    printf("%lld\n", ((qpow(2, n * k, Mod) - rest) + Mod) % Mod);
  }
  int main() {
    GetFrac(frac, 1, 200000, Mod);
    scanf("%lld", &T);
    while (T--) {
      solve();
    }
    return 0;
  }
  