# Makoto and a Blackboard

Makoto has a big blackboard with a positive integer $n$ written on it. He will perform the following action exactly $k$ times:

Suppose the number currently written on the blackboard is $v$. He will randomly pick one of the divisors of $v$ (possibly $1$ and $v$) and replace $v$ with this divisor. As Makoto uses his famous random number generator (RNG) and as he always uses $58$ as his generator seed, each divisor is guaranteed to be chosen with equal probability.

He now wonders what is the expected value of the number written on the blackboard after $k$ steps.

It can be shown that this value can be represented as $\frac{P}{Q}$ where $P$ and $Q$ are coprime integers and $Q \not\equiv 0 \pmod{10^9+7}$. Print the value of $P \cdot Q^{-1}$ modulo $10^9+7$.

Input

The only line of the input contains two integers $n$ and $k$ ($1 \leq n \leq 10^{15}$, $1 \leq k \leq 10^4$).

Output

Print a single integer — the expected value of the number on the blackboard after $k$ steps as $P \cdot Q^{-1} \pmod{10^9+7}$ for $P$, $Q$ defined above.

Examples

input

6 1


output

3


input

6 2


output

875000008


input

60 5


output

237178099


Note

In the first example, after one step, the number written on the blackboard is $1$, $2$, $3$ or $6$ — each occurring with equal probability. Hence, the answer is $\frac{1+2+3+6}{4}=3$.

In the second example, the answer is equal to $1 \cdot \frac{9}{16}+2 \cdot \frac{3}{16}+3 \cdot \frac{3}{16}+6 \cdot \frac{1}{16}=\frac{15}{8}$.

Solution:

#include <bits/stdc++.h>

using namespace std;

const int md = (int) 1e9 + 7;

inline void add(int &a, int b) {
a += b;
if (a >= md) a -= md;
}

inline void sub(int &a, int b) {
a -= b;
if (a < 0) a += md;
}

inline int mul(int a, int b) {
return (int) ((long long) a * b % md);
}

inline int power(int a, long long b) {
int res = 1;
while (b > 0) {
if (b & 1) {
res = mul(res, a);
}
a = mul(a, a);
b >>= 1;
}
return res;
}

inline int inv(int a) {
a %= md;
if (a < 0) a += md;
int b = md, u = 0, v = 1;
while (a) {
int t = b / a;
b -= t * a; swap(a, b);
u -= t * v; swap(u, v);
}
assert(b == 1);
if (u < 0) u += md;
return u;
}

int main() {
ios::sync_with_stdio(false);
cin.tie(0);
long long n;
int k;
cin >> n >> k;
vector<pair<long long, int>> f;
for (long long i = 2; i * i <= n; i++) {
if (n % i == 0) {
int cc = 0;
while (n % i == 0) {
n /= i;
cc++;
}
f.emplace_back(i, cc);
}
}
if (n > 1) {
f.emplace_back(n, 1);
}
int ans = 1;
for (auto& p : f) {
int cc = p.second;
vector<vector<int>> dp(k + 1, vector<int>(cc + 1, 0));
dp[cc] = 1;
for (int i = 0; i < k; i++) {
for (int j = 0; j <= cc; j++) {
int cur = mul(dp[i][j], inv(j + 1));
for (int t = 0; t <= j; t++) {
}
}
}
int x = 1;
int cur = 0;
for (int j = 0; j <= cc; j++) {