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;
long long pow(long long x, long long y) {
  long long ans = 1;
  while (y > 0) {
    if (y & 1) ans = (ans * x);
    y = y >> 1;
    x = (x * x);
  }
  return ans;
}
long long modpow(long long x, long long y, long long p) {
  long long ans = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) ans = (ans * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return ans;
}
long long xxx, yyy, ggg;
void extendedEuclid(long long a, long long b) {
  if (b == 0) {
    ggg = a;
    xxx = 1;
    yyy = 0;
  } else {
    extendedEuclid(b, a % b);
    long long temp = xxx;
    xxx = yyy;
    yyy = temp - (a / b) * yyy;
  }
}
long long modinverse(long long a, long long p) {
  extendedEuclid(a, p);
  if (ggg == 1 && p != 1)
    return (xxx + p) % p;
  else
    return -1;
}
long long add(long long a, long long b, long long m) { return (a + b) % m; }
long long sub(long long a, long long b, long long m) { return (a - b + m) % m; }
long long mult(long long a, long long b, long long m) { return (a * b) % m; }
long long div(long long a, long long b, long long m) {
  return (a * modinverse(b, m)) % m;
}
vector<long long> factors(long long n) {
  vector<long long> fact;
  for (long long i = 2; i * i <= n; i++) {
    if (n % i == 0) fact.push_back(i);
    while (n % i == 0) n = n / i;
  }
  if (n != 1) fact.push_back(n);
  return fact;
}
bool isOverflow(long long a, long long b) {
  long long val1 = log10(a);
  long long val2 = log10(b);
  if (val1 + val2 > 18)
    return true;
  else
    return false;
}
long long solve() {
  long long x, n;
  cin >> x >> n;
  vector<long long> fact = factors(x);
  long long ans = 1;
  for (long long i = 0; i < fact.size(); i++) {
    long long val = fact[i];
    long long cnt = 0;
    while (n / val > 0) {
      cnt = (cnt + (n / val)) % (1000000007 - 1);
      if (isOverflow(val, fact[i]) == true) break;
      val = val * fact[i];
    }
    ans = (ans * modpow(fact[i], cnt, 1000000007)) % 1000000007;
  }
  cout << ans;
  return 0;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  ;
  long long t = 1;
  while (t-- != 0) {
    long long stat = solve();
  }
  return 0;
}
