We'll define S(n) for positive integer n as follows: the number of the n's digits in the decimal base. For example, S(893) = 3, S(114514) = 6.

You want to make a consecutive integer sequence starting from number m (m, m + 1, ...). But you need to pay S(n)·k to add the number n to the sequence.

You can spend a cost up to w, and you want to make the sequence as long as possible. Write a program that tells sequence's maximum length.

Input

The first line contains three integers w (1 ≤ w ≤ 1016), m (1 ≤ m ≤ 1016), k (1 ≤ k ≤ 109).

Please, do not write the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.

Output

The first line should contain a single integer — the answer to the problem.

Examples

Input

9 1 1


Output

9


Input

77 7 7


Output

7


Input

114 5 14


Output

6


Input

1 1 2


Output

0
#include <iostream>
#using namespace std;
#typedef unsigned long long int lli;

#lli w, m, k, lo = 1, hi = 1e18, mid, ans;
w, m, k = map(int, input().split())

'''
lli f(lli x){
    lli p = 1, njir = 1, lol = 0;

    while(10 * p <= x){
        lol = lol + 9 * njir * p;
        njir++;
        p = p * 10;
    }

    return (lol + njir * (x - p + 1)) * k;
}
'''

def f(x):
    p = 1
    njir = 1
    lol = 0

    while 10 * p <= x:
        lol = lol + 9 * njir * p
        njir += 1
        p = p * 10

    return (lol + njir * (x - p + 1)) * k

'''
int main(){
    cin >> w >> m >> k;
    cout << "f(" << m << ") = " << f(m) << '\n';

    while(lo <= hi){
        mid = (lo + hi)/2;
        cout << mid << '\n';

        if(f(mid) <= w + f(m - 1)){
            ans = mid;
            lo = mid + 1;
        }
        else{
            hi = mid - 1;
        }
    }

    cout << ans - m + 1 << '\n';
    return 0;
}
'''

lo = 1
hi = 1
ans = 0
while len(str(hi)) <= 30:
    hi = hi * 10

while lo <= hi:
    mid = (lo + hi)//2

    if f(mid) <= w + f(m - 1):
        ans = mid
        lo = mid + 1
    else:
        hi = mid - 1

print(ans - m + 1)
