Special Offer! Super Price 999 Bourles!

Description

Polycarpus is an amateur businessman. Recently he was surprised to find out that the market for paper scissors is completely free! Without further ado, Polycarpus decided to start producing and selling such scissors.

Polycaprus calculated that the optimal celling price for such scissors would be p bourles. However, he read somewhere that customers are attracted by prices that say something like "Special Offer! Super price 999 bourles!". So Polycarpus decided to lower the price a little if it leads to the desired effect.

Polycarpus agrees to lower the price by no more than d bourles so that the number of nines at the end of the resulting price is maximum. If there are several ways to do it, he chooses the maximum possible price.

Note, Polycarpus counts only the trailing nines in a price.

Input

The first line contains two integers p and d (1 ≤ p ≤ 10¹⁸; 0 ≤ d < p) — the initial price of scissors and the maximum possible price reduction.

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

Output

Print the required price — the maximum price that ends with the largest number of nines and that is less than p by no more than d.

The required number shouldn't have leading zeroes.

Sample Input

Input

1029 102

Output

999

Input

27191 17

Output

27189

题目意思:人们通常对最后带有99的价格比较动心，于是这一个业余商人发现这个商机，决定打折，使价格接近带有99的数，但是又不能为了得到带有最多9的数而减价太多，
所以有一个限度d，减价幅度不能超过d
解题思路：我们可以换一个想法使最后得到的价格的结尾0最多，然后只要减去一个1就能得到最多的9

#include<stdio.h>
int main()
{
    long long int p,d,ans,t;
    while(scanf("%lld%lld",&p,&d)!=EOF)
    {
        t=10;
        p=p+1;
        ans=p;
        while(1)
        {
            if(p%t>d)
            {
                break;
            }
            ans=p-p%t;
            t=t*10;
        }
        printf("%lld
",ans-1);
    }
   return 0;
}