CF1228C.Primes and Multiplication

普及/提高-

通过率:0%

AC君温馨提醒

该题目为【codeforces】题库的题目,您提交的代码将被提交至codeforces进行远程评测,并由ACGO抓取测评结果后进行展示。由于远程测评的测评机由其他平台提供,我们无法保证该服务的稳定性,若提交后无反应,请等待一段时间后再进行重试。

题目描述

Let's introduce some definitions that will be needed later.

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

Let g(x,p)g(x, p) be the maximum possible integer pkp^k where kk is an integer such that xx is divisible by pkp^k . For example:

  • g(45,3)=9g(45, 3) = 9 ( 4545 is divisible by 32=93^2=9 but not divisible by 33=273^3=27 ),
  • g(63,7)=7g(63, 7) = 7 ( 6363 is divisible by 71=77^1=7 but not divisible by 72=497^2=49 ).

Let f(x,y)f(x, y) be the product of g(y,p)g(y, p) for all pp in prime(x)prime(x) . For example:

  • f(30,70)=g(70,2)g(70,3)g(70,5)=213051=10f(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)g(63,5)g(63,7)=325071=63f(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 xx and nn . Calculate f(x,1)f(x,2)f(x,n)mod(109+7)f(x, 1) \cdot f(x, 2) \cdot \ldots \cdot f(x, n) \bmod{(10^{9} + 7)} .

输入格式

The only line contains integers xx and nn ( 2x1092 \le x \le 10^{9} , 1n10181 \le n \le 10^{18} ) — the numbers used in formula.

输出格式

Print the answer.

输入输出样例

  • 输入#1

    10 2
    

    输出#1

    2
    
  • 输入#2

    20190929 1605
    

    输出#2

    363165664
    
  • 输入#3

    947 987654321987654321
    

    输出#3

    593574252
    

说明/提示

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

In the second example, actual value of formula is approximately 1.597101711.597 \cdot 10^{171} . Make sure you print the answer modulo (109+7)(10^{9} + 7) .

In the third example, be careful about overflow issue.

首页