B. Special Offer! Super Price 999 Bourles!

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 test(s)

input

1029 102

output

999

input

27191 17

output

27189

解题说明：题意就是一个物品有一个价格，这个价格可以最多降低d，求在所下降价格不超过d的情况下，能够使价格有最多的9且价格最高。思路是从价格的最后一位开始往前遍历，每次查找最后n位与n个9组成的数之间的差值，如果这个差值在范围内就接受。

#include <iostream>#include<algorithm>#include<cstdio>#include<cmath>#include<cstring>#include<cstdlib>using namespace std;int main(){    long long p, d, i, m = 0;    scanf("%lld%lld", &p, &d);    if (p < 10){printf("%lld\n", p);}else {         for (i = 10 ; i <= p ; i *= 10) {             if (p % i + 1 == i) {continue; } if (p % i + 1 <= d)             { m = p % i + 1; } else { break; }         }         printf("%lld\n", p - m);    }    return 0;}