Codeforces Round #400 (Div. 1 + Div. 2, combined) E. The Holmes Children

E. The Holmes Children

time limit per test

 2 seconds

memory limit per test

 256 megabytes

input

 standard input

output

 standard output

The Holmes children are fighting over who amongst them is the cleverest.

Mycroft asked Sherlock and Eurus to find value of f(n), where f(1) = 1 and for n ≥ 2, f(n) is the number of distinct ordered positive integer pairs (x, y) that satisfy x + y = n and gcd(x, y) = 1. The integer gcd(a, b) is the greatest common divisor of a and b.

Sherlock said that solving this was child's play and asked Mycroft to instead get the value of . Summation is done over all positive integers d that divide n.

Eurus was quietly observing all this and finally came up with her problem to astonish both Sherlock and Mycroft.

She defined a k-composite function Fk(n) recursively as follows:

She wants them to tell the value of Fk(n) modulo 1000000007.

Input

A single line of input contains two space separated integers n (1 ≤ n ≤ 1012) and k (1 ≤ k ≤ 1012) indicating that Eurus asks Sherlock and Mycroft to find the value of Fk(n) modulo 1000000007.

Output

Output a single integer — the value of Fk(n) modulo 1000000007.

Examples

input

7 1

output

6

input

10 2

output

4

Note

In the first case, there are 6 distinct ordered pairs (1, 6), (2, 5), (3, 4), (4, 3), (5, 2) and (6, 1) satisfying x + y = 7 and gcd(x, y) = 1. Hence, f(7) = 6. So, F1(7) = f(g(7)) = f(f(7) + f(1)) = f(6 + 1) = f(7) = 6.

#include <bits/stdc++.h>using namespace std;const int MOD = 1000000007;long long Euler(long long x) {       long long ans = x, i;       for(i = 2; i*i <= x; ++i) {        if(x%i == 0){            ans = ans - ans/i;            while(x%i == 0)x /= i;        }       }       if(x > 1) {        ans = ans - ans / x;       }       return ans;}int main() {        ios::sync_with_stdio(false);        long long n, k;        cin >> n >> k;        while(k>0 && n>1) {                if(k & 1) n = Euler(n);                k--;        }        cout << n % MOD << endl;        return 0;}