Let's introduce some definitions that will be needed later.

Let prime(x) be the set of prime divisors of x. For example, prime(140) = \{ 2, 5, 7 \}, prime(169) = \{ 13 \}.

Let g(x, p) be the maximum possible integer p^k where k is an integer such that x is divisible by p^k. For example:

  * g(45, 3) = 9 (45 is divisible by 3^2=9 but not divisible by 3^3=27), 
  * g(63, 7) = 7 (63 is divisible by 7^1=7 but not divisible by 7^2=49). 



Let f(x, y) be the product of g(y, p) for all p in prime(x). For example:

  * f(30, 70) = g(70, 2) ⋅ g(70, 3) ⋅ g(70, 5) = 2^1 ⋅ 3^0 ⋅ 5^1 = 10, 
  * f(525, 63) = g(63, 3) ⋅ g(63, 5) ⋅ g(63, 7) = 3^2 ⋅ 5^0 ⋅ 7^1 = 63. 



You have integers x and n. Calculate f(x, 1) ⋅ f(x, 2) ⋅ … ⋅ f(x, n) mod{(10^{9} + 7)}.

Input

The only line contains integers x and n (2 ≤ x ≤ 10^{9}, 1 ≤ n ≤ 10^{18}) — the numbers used in formula.

Output

Print the answer.

Examples

Input


10 2


Output


2


Input


20190929 1605


Output


363165664


Input


947 987654321987654321


Output


593574252

Note

In the first example, f(10, 1) = g(1, 2) ⋅ g(1, 5) = 1, f(10, 2) = g(2, 2) ⋅ g(2, 5) = 2.

In the second example, actual value of formula is approximately 1.597 ⋅ 10^{171}. Make sure you print the answer modulo (10^{9} + 7).

In the third example, be careful about overflow issue.
#include <bits/stdc++.h>
using namespace std;
unsigned long long mod = 1e9 + 7;
long long MOD = 998244353;
unsigned long long ncr(long long n, long long k) {
  if (k > n) return 1ll * 0;
  unsigned long long res = 1;
  if (k > n - k) k = n - k;
  for (long long i = 0; i < k; ++i) {
    res *= (n - i);
    res /= (i + 1);
    res = res;
  }
  return res;
}
bool prime(unsigned long long num) {
  bool flag = true;
  for (unsigned long long i = 2; i * i <= num; i++) {
    if (num % i == 0) {
      flag = false;
      break;
    }
  }
  return flag;
}
long long ipow(long long base, long long exp) {
  long long result = 1;
  for (;;) {
    if (exp & 1) {
      result *= base;
      result = result % mod;
    }
    exp >>= 1;
    if (!exp) break;
    base *= base;
    base = base % mod;
  }
  return result % mod;
}
long long abs(long long a, long long b) {
  if ((a - b) > 0)
    return (a - b);
  else
    return (b - a);
}
long long add(long long a, long long b) {
  a += b;
  if (a >= MOD)
    a -= MOD;
  else if (a < 0)
    a += MOD;
  return a;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long der(long long k) {
  long long res = 0;
  long long fact = 1;
  for (long long i = 2; i <= k; i++) fact *= i;
  long long s = 1;
  long long j = 1;
  for (long long i = 2; i <= k; i++) {
    j = j * i;
    long long first = (fact / j) * s;
    res += first;
    s = s * (-1);
  }
  return res;
}
int n, c;
vector<long long> pr;
void _prime(long long n) {
  for (int i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      while (n % i == 0) n = n / i;
      pr.push_back(i);
    }
  }
  if (n != 1) pr.push_back(n);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long x, n;
  cin >> x >> n;
  _prime(x);
  long long ans = 1, tmp;
  tmp = LLONG_MAX;
  for (int i = 0; i < pr.size(); i++) {
    long long tr = n / pr[i];
    long long rf = pr[i];
    long long po = 0;
    long long mxp = (log2(n)) / (log2(rf));
    long long trf = 1;
    for (int j = 0; j < mxp + 1; j++) trf = trf * pr[i];
    if (trf == n) mxp++;
    for (int j = 0; j < mxp; j++) {
      po += n / rf;
      rf = rf * pr[i];
      if (rf < 0) break;
    }
    long long nsum = ipow(pr[i], po);
    ans = (ans * nsum) % mod;
  }
  cout << ans;
}
