CF1097D.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 $10^9+7$.

诚司有一块大黑板，上面写有一个正整数 $n$。他将恰好执行以下操作 $k$ 次：

假设当前黑板上写的数为 $v$。他将随机选取 $v$ 的一个正因数（包括 $1$ 和 $v$ 本身），并用该因数替换 $v$。由于诚司使用他著名的随机数生成器（RNG），且始终以 $58$ 作为生成器种子，因此每个因数被选中的概率均等。

他现在想知道：经过 $k$ 步操作后，黑板上所写数字的期望值是多少？

可以证明，该期望值可表示为最简分数 $\frac{P}{Q}$，其中 $P$ 与 $Q$ 互质，且 $Q \not\equiv 0 \pmod{10^9+7}$。请输出 $P \cdot Q^{-1} \bmod (10^9+7)$ 的值。

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

输入仅包含一行，其中有两个整数 $n$ 和 $k$（$1 \leq n \leq 10^{15}$，$1 \leq k \leq 10^4$）。

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.

输出一个整数——即经过 $k$ 步后黑板上数字的期望值，表示为 $P \cdot Q^{-1} \pmod{10^9+7}$，其中 $P$、$Q$ 如上所定义。

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 $1 \cdot \frac{9}{16}+2 \cdot \frac{3}{16}+3 \cdot \frac{3}{16}+6 \cdot \frac{1}{16}=\frac{15}{8}$.

在第一个例子中，经过一步操作后，黑板上写的数字为 $1$、$2$、$3$ 或 $6$——每种结果出现的概率相等。因此，答案为 $\frac{1+2+3+6}{4}=3$。

在第二个例子中，答案等于 $1 \cdot \frac{9}{16}+2 \cdot \frac{3}{16}+3 \cdot \frac{3}{16}+6 \cdot \frac{1}{16}=\frac{15}{8}$。