Codeforces Round 589 (Div. 2)
C
Primes and Multiplication
Codeforces
1700
原题
← BD →
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Let's introduce some definitions that will be needed later.

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Let $prime(x)$ be the set of prime divisors of $x$. For example, $prime(140) = \{ 2, 5, 7 \}$, $prime(169) = \{ 13 \}$.

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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:

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  • $g(45, 3) = 9$ ($45$ 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 $7^2=49$).
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Let $f(x, y)$ be the product of $g(y, p)$ for all $p$ in $prime(x)$. For example:

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

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

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Input

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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.

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Output

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Print the answer.

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Note

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

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

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In the third example, be careful about overflow issue.

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样例

Input #1
10 2
Output #1
2
InputCopy OutputCopy
Input #2
20190929 1605
Output #2
363165664
InputCopy OutputCopy
Input #3
947 987654321987654321
Output #3
593574252
InputCopy OutputCopy
题目信息
题号
1228C
时间限制
1 s
内存限制
256 MB
标签
数学
数论

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