#include <bits/stdc++.h>
namespace IO {
const int SIZE = (1 << 20) + 1;
char ibuf[SIZE], *iS, *iT, obuf[SIZE], *oS = obuf, *oT = obuf + SIZE - 1;
char _st[55];
int _qr = 0;
inline char gc() {
  return (iS == iT ? iT = (iS = ibuf) + fread(ibuf, 1, SIZE, stdin),
          (iS == iT ? EOF : *iS++) : *iS++);
}
inline void qread() {}
template <class T1, class... T2>
inline void qread(T1 &IEE, T2 &...ls) {
  register T1 __ = 0, ___ = 1;
  register char ch;
  while (!isdigit(ch = gc())) ___ = (ch == '-') ? -___ : ___;
  do {
    __ = (__ << 1) + (__ << 3) + (ch ^ 48);
  } while (isdigit(ch = gc()));
  __ *= ___;
  IEE = __;
  qread(ls...);
  return;
}
template <class T>
inline void QreadArr(T Begin, T End) {
  while (Begin != End) {
    qread(*Begin);
    ++Begin;
  }
}
inline void flush() {
  fwrite(obuf, 1, oS - obuf, stdout);
  oS = obuf;
  return;
}
inline void putc_(char _x) {
  *oS++ = _x;
  if (oS == oT) flush();
}
inline void qwrite() {}
template <class T1, class... T2>
inline void qwrite(T1 IEE, T2... ls) {
  if (!IEE) putc_('0');
  if (IEE < 0) putc_('-'), IEE = -IEE;
  while (IEE) _st[++_qr] = IEE % 10 + '0', IEE /= 10;
  while (_qr) putc_(_st[_qr--]);
  qwrite(ls...);
  return;
}
template <class T>
inline void WriteArr(T Begin, T End) {
  const char Kg = ' ', Edl = '\n';
  while (Begin != End) {
    qwrite(*Begin);
    ++Begin;
    putc_(Kg);
  }
  putc_(Edl);
}
struct Flusher_ {
  ~Flusher_() { flush(); }
} io_flusher;
}  // namespace IO
using namespace IO;
using namespace std;
mt19937_64 rnd(chrono::steady_clock::now().time_since_epoch().count());
long long My_Rand(long long Mod) { return (unsigned long long)(rnd()) % Mod; }
template <class T1, class T2>
T1 Min(const T1 x, const T2 y) {
  return x > y ? y : x;
}
template <class T1, class T2>
T1 Max(const T1 x, const T2 y) {
  return x > y ? x : y;
}
template <class T1, class T2>
void To_max(T1 &x, const T2 y) {
  x < y ? x = y : x = x;
}
template <class T1, class T2>
void To_min(T1 &x, const T2 y) {
  x > y ? x = y : x = x;
}
inline long long qmul(long long x, long long y, long long p) {
  long long l = x * (y >> 25) % p * (1 << 25) % p;
  long long r = x * (y & ((1 << 25) - 1)) % p;
  return (l + r) % p;
}
inline long long qpow(long long n, long long base, long long mod = 1e18) {
  long long ret = 1;
  while (n) {
    if (n & 1) ret = ret * base % mod;
    base = base * base % mod;
    n >>= 1;
  }
  return ret % mod;
}
inline long long gcd(long long x, long long y) {
  return !y ? x : gcd(y, x % y);
}
inline long long lcm(long long x, long long y) { return x / gcd(x, y) * y; }
void InitHash(unsigned long long &base, int &prime, unsigned long long &mod) {
  base = 131, prime = 233317, mod = 212370440130137957ll;
}
const int N = 2e5 + 5;
const int p = 1e9 + 7;
long long inv[N], fac[N], Inv[N];
void init() {
  inv[0] = fac[0] = Inv[0] = inv[1] = fac[1] = Inv[1] = 1;
  for (int i = (2); i <= ((int)(2e5)); ++i) {
    fac[i] = fac[i - 1] * i % p;
    inv[i] = p - (p / i) * inv[p % i] % p;
    Inv[i] = Inv[i - 1] * inv[i] % p;
  }
}
long long C(int n, int m) { return ((fac[n] * Inv[n - m]) % p) * Inv[m] % p; }
int main() {
  init();
  int T;
  qread(T);
  while (T--) {
    int n, k;
    qread(n, k);
    long long ans = 0;
    if (n & 1) {
      ans = 1;
      for (int i = 0; i <= n; i += 2) (ans += C(n, i)) %= p;
    } else
      for (int i = 0; i < n; i += 2) (ans += C(n, i)) %= p;
    long long tmp;
    if (n % 2 == 0) tmp = ans;
    ans = qpow(k, ans, p);
    if (n % 2 == 0) {
      long long Pow = qpow(n, 2, p);
      for (int i = (1); i <= (k); ++i)
        (ans += qpow(i - 1, tmp, p) * qpow(k - i, Pow, p) % p) %= p;
    }
    printf("%lld\n", ans);
  }
  return 0;
}

### Prompt

Your challenge is to write 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>
namespace IO {
const int SIZE = (1 << 20) + 1;
char ibuf[SIZE], *iS, *iT, obuf[SIZE], *oS = obuf, *oT = obuf + SIZE - 1;
char _st[55];
int _qr = 0;
inline char gc() {
  return (iS == iT ? iT = (iS = ibuf) + fread(ibuf, 1, SIZE, stdin),
          (iS == iT ? EOF : *iS++) : *iS++);
}
inline void qread() {}
template <class T1, class... T2>
inline void qread(T1 &IEE, T2 &...ls) {
  register T1 __ = 0, ___ = 1;
  register char ch;
  while (!isdigit(ch = gc())) ___ = (ch == '-') ? -___ : ___;
  do {
    __ = (__ << 1) + (__ << 3) + (ch ^ 48);
  } while (isdigit(ch = gc()));
  __ *= ___;
  IEE = __;
  qread(ls...);
  return;
}
template <class T>
inline void QreadArr(T Begin, T End) {
  while (Begin != End) {
    qread(*Begin);
    ++Begin;
  }
}
inline void flush() {
  fwrite(obuf, 1, oS - obuf, stdout);
  oS = obuf;
  return;
}
inline void putc_(char _x) {
  *oS++ = _x;
  if (oS == oT) flush();
}
inline void qwrite() {}
template <class T1, class... T2>
inline void qwrite(T1 IEE, T2... ls) {
  if (!IEE) putc_('0');
  if (IEE < 0) putc_('-'), IEE = -IEE;
  while (IEE) _st[++_qr] = IEE % 10 + '0', IEE /= 10;
  while (_qr) putc_(_st[_qr--]);
  qwrite(ls...);
  return;
}
template <class T>
inline void WriteArr(T Begin, T End) {
  const char Kg = ' ', Edl = '\n';
  while (Begin != End) {
    qwrite(*Begin);
    ++Begin;
    putc_(Kg);
  }
  putc_(Edl);
}
struct Flusher_ {
  ~Flusher_() { flush(); }
} io_flusher;
}  // namespace IO
using namespace IO;
using namespace std;
mt19937_64 rnd(chrono::steady_clock::now().time_since_epoch().count());
long long My_Rand(long long Mod) { return (unsigned long long)(rnd()) % Mod; }
template <class T1, class T2>
T1 Min(const T1 x, const T2 y) {
  return x > y ? y : x;
}
template <class T1, class T2>
T1 Max(const T1 x, const T2 y) {
  return x > y ? x : y;
}
template <class T1, class T2>
void To_max(T1 &x, const T2 y) {
  x < y ? x = y : x = x;
}
template <class T1, class T2>
void To_min(T1 &x, const T2 y) {
  x > y ? x = y : x = x;
}
inline long long qmul(long long x, long long y, long long p) {
  long long l = x * (y >> 25) % p * (1 << 25) % p;
  long long r = x * (y & ((1 << 25) - 1)) % p;
  return (l + r) % p;
}
inline long long qpow(long long n, long long base, long long mod = 1e18) {
  long long ret = 1;
  while (n) {
    if (n & 1) ret = ret * base % mod;
    base = base * base % mod;
    n >>= 1;
  }
  return ret % mod;
}
inline long long gcd(long long x, long long y) {
  return !y ? x : gcd(y, x % y);
}
inline long long lcm(long long x, long long y) { return x / gcd(x, y) * y; }
void InitHash(unsigned long long &base, int &prime, unsigned long long &mod) {
  base = 131, prime = 233317, mod = 212370440130137957ll;
}
const int N = 2e5 + 5;
const int p = 1e9 + 7;
long long inv[N], fac[N], Inv[N];
void init() {
  inv[0] = fac[0] = Inv[0] = inv[1] = fac[1] = Inv[1] = 1;
  for (int i = (2); i <= ((int)(2e5)); ++i) {
    fac[i] = fac[i - 1] * i % p;
    inv[i] = p - (p / i) * inv[p % i] % p;
    Inv[i] = Inv[i - 1] * inv[i] % p;
  }
}
long long C(int n, int m) { return ((fac[n] * Inv[n - m]) % p) * Inv[m] % p; }
int main() {
  init();
  int T;
  qread(T);
  while (T--) {
    int n, k;
    qread(n, k);
    long long ans = 0;
    if (n & 1) {
      ans = 1;
      for (int i = 0; i <= n; i += 2) (ans += C(n, i)) %= p;
    } else
      for (int i = 0; i < n; i += 2) (ans += C(n, i)) %= p;
    long long tmp;
    if (n % 2 == 0) tmp = ans;
    ans = qpow(k, ans, p);
    if (n % 2 == 0) {
      long long Pow = qpow(n, 2, p);
      for (int i = (1); i <= (k); ++i)
        (ans += qpow(i - 1, tmp, p) * qpow(k - i, Pow, p) % p) %= p;
    }
    printf("%lld\n", ans);
  }
  return 0;
}

```