Primes and Multiplication

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:

• $f(30, 70) = g(70, 2) \cdot g(70, 3) \cdot g(70, 5) = 2^1 \cdot 3^0 \cdot 5^1 = 10$,
• $f(525, 63) = g(63, 3) \cdot g(63, 5) \cdot g(63, 7) = 3^2 \cdot 5^0 \cdot 7^1 = 63$.

You have integers $x$ and $n$. Calculate $f(x, 1) \cdot f(x, 2) \cdot \ldots \cdot f(x, n) \bmod{(10^{9} + 7)}$.

The only line contains integers $x$ and $n$ ($2 \le x \le 10^{9}$, $1 \le n \le 10^{18}$) --- the numbers used in formula.

Print the answer.

In the first example, $f(10, 1) = g(1, 2) \cdot g(1, 5) = 1$, $f(10, 2) = g(2, 2) \cdot g(2, 5) = 2$.

In the second example, actual value of formula is approximately $1.597 \cdot 10^{171}$. Make sure you print the answer modulo $(10^{9} + 7)$.

In the third example, be careful about overflow issue.