#include <bits/stdc++.h>
#pragma GCC optimize("O3")
#pragma GCC optimize("-Ofast", "-funroll-all-loops", "-ffast-math")
#pragma GCC target("avx2,fma")
#pragma GCC optimize("unroll-loops")
using namespace std;
const long long Linf = 0x7f7f7f7f7f7f7f7f;
const long long Inf = 0x3f3f3f3f;
const long long MAXN = 1e6;
const long long P = 1e9 + 7;
long long qpow(long long a, long long b) {
  long long res = 1;
  a %= P;
  while (b) {
    if (b & 1) res = res * a % P;
    a = 1ll * a * a % P;
    b >>= 1;
  }
  return res % P;
}
long long fac[MAXN], inv[MAXN];
long long C(long long n, long long m) {
  return ((1ll * fac[n] * inv[n - m] % P) * inv[m]) % P;
}
long long n, m, dp[MAXN][2];
void doit() {
  cin >> n >> m;
  long long res1 = 0, res0 = 0;
  for (long long i = 0; i < n; i++) {
    if (i & 1)
      (res1 += C(n, i)) %= P;
    else
      (res0 += C(n, i)) %= P;
  }
  long long ans = 0;
  if (n & 1) {
    (ans = qpow(1 + res0, m)) %= P;
  } else {
    long long mul = 1;
    ans = 0;
    for (long long i = m; i >= 1; i--, (mul *= res0) %= P) {
      long long tp = qpow(qpow(2, i - 1), n);
      (ans += mul * tp % P) %= P;
    }
    (ans += mul) %= P;
  }
  cout << ans % P << '\n';
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  fac[0] = 1;
  inv[0] = 1;
  for (long long i = 1; i < MAXN; i++) fac[i] = 1ll * fac[i - 1] * i % P;
  for (long long i = 1; i < MAXN; i++) inv[i] = qpow(fac[i], P - 2);
  long long _;
  cin >> _;
  while (_--) doit();
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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].

### Response

```cpp
#include <bits/stdc++.h>
#pragma GCC optimize("O3")
#pragma GCC optimize("-Ofast", "-funroll-all-loops", "-ffast-math")
#pragma GCC target("avx2,fma")
#pragma GCC optimize("unroll-loops")
using namespace std;
const long long Linf = 0x7f7f7f7f7f7f7f7f;
const long long Inf = 0x3f3f3f3f;
const long long MAXN = 1e6;
const long long P = 1e9 + 7;
long long qpow(long long a, long long b) {
  long long res = 1;
  a %= P;
  while (b) {
    if (b & 1) res = res * a % P;
    a = 1ll * a * a % P;
    b >>= 1;
  }
  return res % P;
}
long long fac[MAXN], inv[MAXN];
long long C(long long n, long long m) {
  return ((1ll * fac[n] * inv[n - m] % P) * inv[m]) % P;
}
long long n, m, dp[MAXN][2];
void doit() {
  cin >> n >> m;
  long long res1 = 0, res0 = 0;
  for (long long i = 0; i < n; i++) {
    if (i & 1)
      (res1 += C(n, i)) %= P;
    else
      (res0 += C(n, i)) %= P;
  }
  long long ans = 0;
  if (n & 1) {
    (ans = qpow(1 + res0, m)) %= P;
  } else {
    long long mul = 1;
    ans = 0;
    for (long long i = m; i >= 1; i--, (mul *= res0) %= P) {
      long long tp = qpow(qpow(2, i - 1), n);
      (ans += mul * tp % P) %= P;
    }
    (ans += mul) %= P;
  }
  cout << ans % P << '\n';
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  fac[0] = 1;
  inv[0] = 1;
  for (long long i = 1; i < MAXN; i++) fac[i] = 1ll * fac[i - 1] * i % P;
  for (long long i = 1; i < MAXN; i++) inv[i] = qpow(fac[i], P - 2);
  long long _;
  cin >> _;
  while (_--) doit();
  return 0;
}

```