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;
const long long int INF = 9223372036854775807;
const long long int mod = (long long int)1e9 + 7;
long long int max3(long long int a, long long int b, long long int c) {
  return max(c, max(a, b));
}
long long int min3(long long int a, long long int b, long long int c) {
  return min(a, min(b, c));
}
long long int power(long long int x, long long int y, long long int mod1) {
  long long int res = 1;
  while (y > 0) {
    if (y & 1) res = (res * x) % mod1;
    y = y >> 1;
    x = (x * x) % mod1;
  }
  return res;
}
long long int logg(long long int a) {
  long long int x = 0;
  while (a > 1) {
    x++;
    a /= 2;
  }
  return x;
}
long long int gcd(long long int a, long long int b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long int abso(long long int x) {
  if (x < 0) {
    return -x;
  }
  return x;
}
long long int ceiling(long long int a, long long int b) {
  if (a % b == 0) {
    return a / b;
  } else {
    return a / b + 1;
  }
}
long long int modinv(long long int x, long long int p) {
  return power(x, p - 2, p);
}
struct point {
  long long int x;
  long long int y;
};
long long int cross(point a, point b) { return a.x * b.y - b.x * a.y; }
bool ccw(point p1, point p2) {
  if (cross(p1, p2) > 0) {
    return true;
  } else {
    return false;
  }
}
void ccw_sort(vector<point> &v) {
  vector<point> v1, v2;
  for (int i = 0; i < v.size(); i++) {
    if (v[i].y > 0) {
      v1.push_back(v[i]);
    } else {
      v2.push_back(v[i]);
    }
  }
  sort(v1.begin(), v1.end(), ccw);
  sort(v2.begin(), v2.end(), ccw);
  for (int i = 0; i < v2.size(); i++) {
    v1.push_back(v2[i]);
  }
  v = v1;
}
int main() {
  long long int x, n;
  cin >> x >> n;
  long long int y = x;
  vector<long long int> pr;
  for (int i = 2; i * i <= y; i++) {
    if (x % i == 0) {
      pr.push_back(i);
      while (x % i == 0) {
        x = x / i;
      }
    }
  }
  if (x > 1) {
    pr.push_back(x);
  }
  long long int ans = 1;
  for (int i = 0; i < pr.size(); i++) {
    long long int prod = 1;
    long long int pw = 0;
    while (prod <= n / pr[i]) {
      prod = prod * pr[i];
      pw += (n / prod);
    }
    ans = ((ans * (power(pr[i], pw, mod))) % mod);
  }
  cout << ans << endl;
}
