Trailing Loves (or L'oeufs?)

The number "zero" is called "love" (or "l'oeuf" to be precise, literally means "egg" in French), for example when denoting the zero score in a game of tennis.

Aki is fond of numbers, especially those with trailing zeros. For example, the number 9200 has two trailing zeros. Aki thinks the more trailing zero digits a number has, the prettier it is.

However, Aki believes, that the number of trailing zeros of a number is not static, but depends on the base (radix) it is represented in. Thus, he considers a few scenarios with some numbers and bases. And now, since the numbers he used become quite bizarre, he asks you to help him to calculate the beauty of these numbers.

Given two integers n and b (in decimal notation), your task is to calculate the number of trailing zero digits in the b-ary (in the base/radix of b) representation of n! (factorial of n).

Input

The only line of the input contains two integers n and b (1≤n≤10^18  ,   2<=b<=10^12

).

Output

Print an only integer — the number of trailing zero digits in the b-ary representation of n!

Examples

Input

6 9

Output

1

Input

38 11

Output

3

Input

5 2

Output

3

Input

5 10

Output

1

Note

In the first example, 6!(10)=720(10)=880(9).

In the third and fourth example, 5!(10)=120(10)=1111000(2).

The representation of the number x in the b-ary base is d1,d2,…,dk if x=d1bk−1+d2bk−2+…+dkb0, where di are integers and 0≤di≤b−1. For example, the number 720 from the first example is represented as 880(9) since 720=8⋅92+8⋅9+0⋅1.

思路：把b分解质因数，然后看对n！献出了多少贡献，即（n！%（a1^k+a2^k......)==0

我们需要去求k，就需要先把b分解，并且记录下它的质因子的指数数，然后用n进行迭代求，然后每次缩小一次指数，最后除本身的指数就ok了，注意minn开的一定要尽可能的大

代码：

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<cmath>

typedef long long ll;

using namespace std;

ll cnt=0;
ll num[4000005];

void primeFactor(ll n) {
	while(n % 2 == 0) {
		num[cnt++]=2;
		n /= 2;
	}
	for(ll i = 3; i <= sqrt(n); i += 2) {
		while(n % i == 0) {
			num[cnt++]=i;
			n /= i;
		}
	}
	if(n > 2)
		num[cnt++]=n;
}
int main() {

	ll n,b;
	ll ans;
	ll sss;
	scanf("%lld%lld",&n,&b);
	primeFactor(b);
	ll s;
	ll ss;
	ll k=1;
	ll minn=999999999999999999;
	for(int t=0; t<cnt; t++) {
		if(num[t]!=num[t+1]) {
			s=0;
			ans=n;
			ss=num[t];

			while(ans>=ss) {
				s+=(ans/ss);
				ans/=ss;
			}
			minn=min(minn,s/k);
			k=1;

		} else {
			k++;
		}
//	    cout<<num[t]<<endl;
	}

	printf("%lld",minn);
	return 0;

}