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 = 1e6 + 10;
const int mod = 1e9 + 7;
int head[N];
int dis[N], ecnt;
int C[4005][4005];
int cat[2005][2005];
struct EDGE {
  int u, v, nxt, w;
} e[N];
void add_edge(int u, int v, int w) {
  e[ecnt].u = u;
  e[ecnt].v = v;
  e[ecnt].w = w;
  e[ecnt].nxt = head[u];
  head[u] = ecnt++;
}
long long m, n, a, b, k, x;
string s;
long long ar[N], br[N];
bool vis[N];
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  if (b == 0) return a;
  return a % b == 0 ? b : gcd(b, a % b);
}
long long qpow(long long x, long long base) {
  long long ans = 1;
  while (x) {
    if (x & 1) ans = (ans * base) % mod;
    x >>= 1;
    base = base * base % mod;
  }
  return ans;
}
vector<int> ed[N];
map<long long, int> dt[N];
vector<int> pm;
long long ans = 1;
void dfs(long long res) {
  for (int i = 0; i < pm.size(); i++) {
    long long tp = res * pm[i];
    if (tp <= n) {
      ans = (ans * (tp % mod)) % mod;
      dfs(tp);
    }
  }
}
int main() {
  scanf("%d %lld", &x, &n);
  for (int i = 2; i * i <= x; i++) {
    if (x % i == 0) pm.push_back(i);
    while (x % i == 0) {
      x /= i;
    }
  }
  if (x > 1) pm.push_back(x);
  for (int i = 0; i < pm.size(); i++) {
    long long k = n;
    long long tm = 0;
    while (pm[i] <= k) {
      tm += k / pm[i];
      k /= pm[i];
    }
    ans = ans * qpow(tm, pm[i]) % mod;
  }
  cout << ans << endl;
  return 0;
}
