Problem statement

There are two integers $ A $ and $ B $. The process of finding $ A --B $ by writing in decimal is shown below.

1. Express $ A $ and $ B $ in decimal. Subscripts are $ 0,1,2, ..., n-1 $ in order from the bottom, $ A = A_ {n-1} A_ {n-2} ... A_ {0} $, $ B = B_ { n-1} B_ {n-2} ... B_ {0} $. $ n $ is the number of digits of $ A $ when expressed in decimal. When the number of digits of $ B $ is less than $ n $, the upper digit is supplemented with $ 0 $.
2. Set $ ​​borrow_ {0} = 0 $.
3. Repeat the following process from $ i = 0 $ to $ n-1 $.
3.1. If $ A_ {i} --borrow_ {i} \ geq B_ {i} $, then $ C_ {i} = A_ {i} --borrow_ {i} --B_ {i} $, $ borrow_ {i + 1 } = 0 $.
3.2. If $ A_ {i} --borrow_ {i} \ lt B_ {i} $, then $ C_ {i} = A_ {i} --borrow_ {i} + 10 --B_ {i} $, $ borrow_ {i Let +1} = 1 $.

The calculation result of $ A --B $ is $ C_ {n-1} C_ {n-2} ... C_ {0} $. However, the top-level consecutive $ 0 $ is removed.

If you forget to carry down, $ borrow_ {i + 1} = 1 $ in the process of 3.2 becomes $ borrow_ {i + 1} = 0 $. Find the maximum value of the calculation result when you forget to carry down the maximum $ K $ times.

Constraint

* $ 1 \ leq B \ lt A \ leq 10 ^ 9 $
* $ 1 \ leq K \ leq 9 $



input

Input follows the following format. All given numbers are integers.




$ A $ $ B $ $ K $

output

Output the maximum value of the calculation result on one line.

Examples

Input

99 98 1


Output

1


Input

100 2 3


Output

198


Input

538 84 1


Output

554


Input

2012 1987 1


Output

1025
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cmath>

#define ll long long

using namespace std;

int digitAt(int arg, int d){
    int ret;

    ret = arg % (int)pow(10, d+1);
    ret /= (int)pow(10, d);
    return ret;
}

int bitcount(int arg){
    int ret=0;
    while(arg > 0){
        if(arg & 1) ret++;
        arg >>= 1;
    }
    return ret;
}

int main(){
    int a,b,k,tmp=0,borrow=0,ta,tb,ret=0;
    cin >> a >> b >> k;

    int n = log10(a) + 1;

    for(int i=0;i<(1 << n);i++){
        if(bitcount(i) <= k){
            tmp = 0;
            borrow = 0;
            for(int j=0;j<n;j++){
                ta = digitAt(a,j);
                tb = digitAt(b,j);
                if(ta - borrow >= tb){
                    tmp += (ta - tb - borrow) * pow(10, j);
                    borrow = 0;
                }else{
                    tmp += (ta - tb - borrow + 10) * pow(10, j);
                    if((1 << j) & i){
                        borrow = 0;
                    }else{
                        borrow = 1;
                    }
                }
            }
            ret = max(ret, tmp);
        }
    }
    cout << ret << endl;
}