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 int N = 2e3 + 5;
const int mod = 1e9 + 7;
const int mod2 = 1e9 + 6;
int r[N];
int c[N];
int mat[N][N];
int add(int a, int b) {
  a += b;
  if (a >= mod) {
    a -= mod;
  }
  return a;
}
int add2(int a, int b) {
  a += b;
  if (a >= mod2) {
    a -= mod2;
  }
  return a;
}
int mul(int a, int b) { return (1LL * (a % mod) * (b % mod)) % mod; }
int mul2(int a, int b) { return (1LL * (a % mod2) * (b % mod2)) % mod2; }
int power(int a, int b) {
  int res = 1;
  while (b) {
    if (b & 1) {
      res = mul(res, a);
    }
    a = mul(a, a);
    b /= 2;
  }
  return res;
}
int invmd(int x) { return power(x, mod - 2); }
vector<int> getPrimes(int x) {
  vector<int> v;
  for (int i = 2; 1LL * i * i <= x; ++i) {
    if (x % i == 0) {
      v.push_back(i);
      while (x % i == 0) {
        x /= i;
      }
    }
  }
  if (x > 1) {
    v.push_back(x);
  }
  return v;
}
int contribution(int p, long long n) {
  int till = 0;
  long long cur = 1;
  do {
    if (cur <= (n / p)) {
      till++;
      cur = 1LL * cur * p;
    } else {
      break;
    }
  } while (1);
  vector<int> cnt(till + 1, 0);
  vector<long long> v(till + 1, 0);
  cur = 1;
  for (int i = 1; i <= till; ++i) {
    cur = 1LL * cur * p;
    v[i] = (n / cur);
  }
  int sum = 0;
  for (int i = till; i >= 1; --i) {
    for (int j = i + 1; j <= till; j++) {
      v[i] -= v[j];
    }
    sum = add2(sum, mul2(i, v[i] % mod2));
  }
  return power(p, sum);
}
int brute(vector<int>& primes, int n) {
  int res = 1;
  for (auto p : primes) {
    int tot = 0;
    for (int i = 1; i <= n; ++i) {
      long long cur = 1;
      long long last = 0;
      int val = i;
      int cnt = -1;
      while ((val % cur) == 0) {
        last = cur;
        cnt++;
        cur = 1LL * cur * p;
      }
      res = mul(res, last);
      tot += cnt;
    }
  }
  return res;
}
int solve(vector<int>& primes, long long n) {
  int res = 1;
  for (auto p : primes) {
    res = mul(res, contribution(p, n));
  }
  return res;
}
void solve() {
  ios_base ::sync_with_stdio(false);
  cin.tie(NULL);
  int x;
  long long n;
  cin >> x >> n;
  vector<int> primes = getPrimes(x);
  int ans = solve(primes, n);
  cout << ans << endl;
}
int main() {
  solve();
  return 0;
}
