#include <bits/stdc++.h>
using namespace std;
template <class c>
struct rge {
  c b, e;
};
template <class c>
rge<c> range(c i, c j) {
  return rge<c>{i, j};
}
template <class c>
auto dud(c* x) -> decltype(cerr << *x, 0);
template <class c>
char dud(...);
struct debug {
  template <class c>
  debug& operator<<(const c&) {
    return *this;
  }
};
vector<char*> tokenizer(const char* args) {
  char* token = new char[111];
  strcpy(token, args);
  token = strtok(token, ", ");
  vector<char*> v({token});
  while (token = strtok(NULL, ", ")) v.push_back(token);
  return reverse(v.begin(), v.end()), v;
}
void debugg(vector<char*> args) { cerr << "\b "; }
template <typename Head, typename... Tail>
void debugg(vector<char*> args, Head H, Tail... T) {
  debug() << " [" << args.back() << ": " << H << "] ";
  args.pop_back();
  debugg(args, T...);
}
const double PI = acos(-1);
const long long mod = 1e9 + 7;
long long bin_pow(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return res;
}
long long n;
vector<long long> dp;
long long sol(long long k) {
  {};
  if (k == 1) {
    if (n % 2 == 0)
      return bin_pow(2, n - 1);
    else
      return (1 + bin_pow(2, n - 1)) % mod;
  }
  if (dp[k] != -1) return dp[k];
  dp[k] = 0;
  if (n % 2 == 0) {
    dp[k] += bin_pow(2, k * n - n);
    dp[k] %= mod;
    dp[k] += (((bin_pow(2, n - 1) - 1 + mod) % mod) * sol(k - 1)) % mod;
    dp[k] %= mod;
  } else {
    dp[k] += sol(k - 1);
    dp[k] %= mod;
    dp[k] += (bin_pow(2, n - 1) * sol(k - 1)) % mod;
    dp[k] %= mod;
  }
  return dp[k];
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  int test = 1;
  cin >> test;
  while (test--) {
    long long k;
    cin >> n >> k;
    dp = vector<long long>(k + 1, -1);
    dp[0] = 1;
    cout << sol(k) << "\n";
  }
  return 0;
}

### Prompt

Your task is to create a Cpp solution to the following problem:

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>
using namespace std;
template <class c>
struct rge {
  c b, e;
};
template <class c>
rge<c> range(c i, c j) {
  return rge<c>{i, j};
}
template <class c>
auto dud(c* x) -> decltype(cerr << *x, 0);
template <class c>
char dud(...);
struct debug {
  template <class c>
  debug& operator<<(const c&) {
    return *this;
  }
};
vector<char*> tokenizer(const char* args) {
  char* token = new char[111];
  strcpy(token, args);
  token = strtok(token, ", ");
  vector<char*> v({token});
  while (token = strtok(NULL, ", ")) v.push_back(token);
  return reverse(v.begin(), v.end()), v;
}
void debugg(vector<char*> args) { cerr << "\b "; }
template <typename Head, typename... Tail>
void debugg(vector<char*> args, Head H, Tail... T) {
  debug() << " [" << args.back() << ": " << H << "] ";
  args.pop_back();
  debugg(args, T...);
}
const double PI = acos(-1);
const long long mod = 1e9 + 7;
long long bin_pow(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return res;
}
long long n;
vector<long long> dp;
long long sol(long long k) {
  {};
  if (k == 1) {
    if (n % 2 == 0)
      return bin_pow(2, n - 1);
    else
      return (1 + bin_pow(2, n - 1)) % mod;
  }
  if (dp[k] != -1) return dp[k];
  dp[k] = 0;
  if (n % 2 == 0) {
    dp[k] += bin_pow(2, k * n - n);
    dp[k] %= mod;
    dp[k] += (((bin_pow(2, n - 1) - 1 + mod) % mod) * sol(k - 1)) % mod;
    dp[k] %= mod;
  } else {
    dp[k] += sol(k - 1);
    dp[k] %= mod;
    dp[k] += (bin_pow(2, n - 1) * sol(k - 1)) % mod;
    dp[k] %= mod;
  }
  return dp[k];
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  int test = 1;
  cin >> test;
  while (test--) {
    long long k;
    cin >> n >> k;
    dp = vector<long long>(k + 1, -1);
    dp[0] = 1;
    cout << sol(k) << "\n";
  }
  return 0;
}

```