Educational Codeforces Round 23 C. Really Big Numbers 暴力

C. Really Big Numbers

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Ivan likes to learn different things about numbers, but he is especially interested in really big numbers. Ivan thinks that a positive integer number x is really big if the difference between x and the sum of its digits (in decimal representation) is not less than s. To prove that these numbers may have different special properties, he wants to know how rare (or not rare) they are — in fact, he needs to calculate the quantity of really big numbers that are not greater than n.

Ivan tried to do the calculations himself, but soon realized that it's too difficult for him. So he asked you to help him in calculations.

Input

The first (and the only) line contains two integers n and s (1 ≤ n, s ≤ 10¹⁸).

Output

Print one integer — the quantity of really big numbers that are not greater than n.

Examples

Input

12 1

Output

3

Input

25 20

Output

0

Input

10 9

Output

1

Note

In the first example numbers 10, 11 and 12 are really big.

In the second example there are no really big numbers that are not greater than 25 (in fact, the first really big number is 30: 30 - 3 ≥ 20).

In the third example 10 is the only really big number (10 - 1 ≥ 9).

题意：找出1-n内所有数字，数字要满足的条件：数字-每位数的和>=s;

思路：每位数的和最多180，所有只需要暴力s，s+180区间，后面小于n的肯定满足条件；

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<set>
#include<map>
#include<bitset>
#include<time.h>
using namespace std;
#define LL long long
#define pi (4*atan(1.0))
#define eps 1e-4
#define bug(x)  cout<<"bug"<<x<<endl;
const int N=3e5+,M=1e5+,inf=,mod=1e9+;
const LL INF=1e18+,MOD=1e9+;

LL sum(LL x)
{
    if(x==)return ;
    return sum(x/)+x%;
}
int main()
{
    LL n,s;
    scanf("%lld%lld",&n,&s);
    LL x=min(n,s);
    LL ans=;
    for(LL i=s;i<=s+;i++)
    {
        if(i<=n&&i-sum(i)>=s)
            ans++;
    }
    ans+=max(0LL,n-s-);
    printf("%lld\n",ans);
    return ;
}