from collections import defaultdict, deque, Counter
from heapq import heapify, heappop, heappush
import math
from copy import deepcopy
from itertools import combinations, permutations, product, combinations_with_replacement
from bisect import bisect_left, bisect_right

import sys

def input():
    return sys.stdin.readline().rstrip()
def getN():
    return int(input())
def getNM():
    return map(int, input().split())
def getList():
    return list(map(int, input().split()))
def getListGraph():
    return list(map(lambda x:int(x) - 1, input().split()))
def getArray(intn):
    return [int(input()) for i in range(intn)]

mod = 10 ** 9 + 7
MOD = 998244353
# sys.setrecursionlimit(10000000)
inf = float('inf')
eps = 10 ** (-15)
dy = [0, 1, 0, -1]
dx = [1, 0, -1, 0]

#############
# Main Code #
#############

def prime_factorize(n):
    divisors = []
    # 27(2 * 2 * 7)の7を出すためにtemp使う
    temp = n
    for i in range(2, int(math.sqrt(n)) + 1):
        if temp % i == 0:
            cnt = 0
            while temp % i == 0:
                cnt += 1
                # 素因数を見つけるたびにtempを割っていく
                temp //= i
            divisors.append([i, cnt])
    if temp != 1:
        divisors.append([temp, 1])
    if divisors == []:
        divisors.append([n, 1])

    return divisors

"""
x = 0なら余りは0
nCkは求められない　どんな数か？
Mの因数はいくつあるか？という問題　分子と分母でMの個数を数えてみる
nCk = n! / (n - r)!r! とすると簡単
8*7*6*5*4*3*2*1/3*2*1 * 5*4*3*2*1

まず素数で割る
"""

N, K, M = getNM()

# 素因数の数を数える
def calc(x, m):
    tar = m
    res = 0
    while tar <= x:
        res += x // tar
        tar *= m
    return res

def count(n, r, m):
    return calc(n, m) - calc(n - r, m) - calc(r, m)

ans = inf
# 各因数について見ていく
for k, v in prime_factorize(M):
    ans = min(ans, count(N, K, k) // v)
print(ans)