#include <bits/stdc++.h>
using namespace std;
long long mult(long long a, long long b, long long p = 1000000007) {
  long long ans = ((a % p) * (b % p)) % p;
  return (ans + p) % p;
}
long long add(long long a, long long b, long long p = 1000000007) {
  long long ans = (a % p + b % p) % p;
  return (ans + p) % p;
}
long long fpow(long long n, long long k, long long p = 1000000007) {
  long long r = 1;
  for (; k; k >>= 1) {
    if (k & 1) r = r * n % p;
    n = n * n % p;
  }
  return r;
}
long long inv(long long a, long long p = 1000000007) {
  return fpow(a, p - 2, p);
}
long long dp[64][int(1e4) + 1];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n, k;
  cin >> n >> k;
  for (int j = 0; j < k + 1; j++) dp[0][j] = 1;
  vector<pair<long long, long long> > primedivs;
  long long dupn = n;
  for (long long i = 2; i * i <= n; ++i) {
    long long c = 0;
    while (dupn % i == 0) {
      dupn /= i;
      c++;
    }
    if (c) primedivs.push_back({i, c});
  }
  if (dupn != 1) primedivs.push_back({dupn, 1});
  long long ans = 1;
  for (auto iter : primedivs) {
    long long p = iter.first, pow = iter.second;
    for (int i = 1; i < pow + 1; i++) dp[i][0] = fpow(p, i);
    for (int i = 1; i < pow + 1; i++) {
      for (int j = 1; j < k + 1; j++) {
        dp[i][j] = add(dp[i][j - 1], mult(dp[i - 1][j], i));
        dp[i][j] = mult(dp[i][j], fpow(i + 1, 1000000007 - 2));
      }
    }
    ans = mult(ans, dp[pow][k]);
  }
  cout << ans << endl;
  return 0;
}
