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;
inline int read() {
  int f = 1, x = 0;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
inline long long readll() {
  long long f = 1, x = 0;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
const int MOD = 1e9 + 7;
long long v[100005], p[100005], cnt;
void sieve() {
  v[1] = 1;
  for (long long i = 2; i <= 100000; ++i) {
    if (!v[i]) p[++cnt] = i;
    for (long long j = 1; j <= cnt && (long long)i * p[j] <= 100000; ++j) {
      v[i * p[j]] = 1;
      if (i % p[j] == 0) break;
    }
  }
}
long long x, px[100005], c;
long long n;
long long pow_mod(long long x, long long i) {
  long long y = 1;
  while (i) {
    if (i & 1LL) y = y * x % MOD;
    x = x * x % MOD;
    i >>= 1;
  }
  return y;
}
bool isp(long long x) {
  for (long long i = 2; (long long)i * i <= (long long)x; ++i)
    if (x % i == 0) return false;
  return true;
}
signed main() {
  sieve();
  x = read();
  n = readll();
  for (long long i = 1; (long long)i * i <= (long long)x; ++i) {
    if (x % i == 0) {
      if (!v[i]) px[++c] = i;
      if (i != x / i) {
        if (x / i > 100000) {
          if (isp(x / i)) px[++c] = x / i;
        } else if (!v[x / i])
          px[++c] = x / i;
      }
    }
  }
  long long ans = 1;
  for (long long i = 1; i <= c; ++i) {
    long long p = px[i], cur = p;
    while (cur <= n) {
      long long num = n / cur;
      ans = ans * pow_mod(p, num) % MOD;
      if (cur > n / p) break;
      cur = cur * p;
      if (cur < 0) break;
    }
  }
  cout << ans << endl;
  return 0;
}
