You have a fraction <image>. You need to find the first occurrence of digit c into decimal notation of the fraction after decimal point.

Input

The first contains three single positive integers a, b, c (1 ≤ a < b ≤ 105, 0 ≤ c ≤ 9).

Output

Print position of the first occurrence of digit c into the fraction. Positions are numbered from 1 after decimal point. It there is no such position, print -1.

Examples

Input

1 2 0


Output

2

Input

2 3 7


Output

-1

Note

The fraction in the first example has the following decimal notation: <image>. The first zero stands on second position.

The fraction in the second example has the following decimal notation: <image>. There is no digit 7 in decimal notation of the fraction. 
import sys,math,io,os,time,itertools,collections
mod=10**9+7
sys.setrecursionlimit(10000)
i=sys.stdin.readline
p=sys.stdout.write
#use sys.stdout.write() (remember to convert to str b4 and concatenate "\n")
global start,end

#binary search
def isin(l,x):
    left=0
    right=len(l)-1
    if x<l[0]:
        return -1
    while left<=right:
        mid=left + (right -left)//2
        if l[mid]==x:
            return mid
        elif l[mid]<x:
            ans=mid
            left=mid+1
        else:
            right=mid-1
    return ans

#is palindrome or not
def ispal(l):
    n=len(l)
    for i in range(n//2+1):
        if l[i]!=l[n-i-1]:
            return False
    return True

#coordinate compression
def ccarray(l):
    d={l[k]:k for k in range(len(l))}
    m=sorted(d)
    return [d[m[k]] for k in range(len(l))]

#checks if prime or not
def is_prime(n):
    if n<=3:
        return n>1
    if n%2==0 or n%3==0:
        return False
    k=5
    while k**2<=n:
        if n%k==0 or n%(k+2)==0:
            return False
        k+=6
    return True

#sieve of eratosthenes
def sieve(n):
    prime=[True for k in range(n+1)]
    p=2
    while p*p<=n:
        if prime[p]==True:
            for k in range(p*p,n+1,p):
                prime[k]=False
        p+=1

def main():
    a,b,c=[int(k) for k in i().split()]
    for k in range(b+1):
        num=(a%b)*10
        digit=num//b
        if digit==c:
            p(str(k+1)+"\n")
            break
        a=num
    else:
        p("-1\n")

t=1
#t=int(i())
start=time.perf_counter()
for _ in range(t):
    main()
end=time.perf_counter()
#print(end-start)