Codeforces Round #427 (Div. 2) B. The number on the board

引子：

A题过于简单导致不敢提交，拖拖拉拉10多分钟还是决定交，太冲动交错了CE一发，我就知道又要错过一次涨分的机会....

B题还是过了，根据题意目测数组大小开1e5，居然蒙对，感觉用vector更好一点...

C题WA第9组，GG思密达....明天起床再补C吧 - -

题目链接：

http://codeforces.com/contest/835/problem/B

B. The number on the board

Some natural number was written on the board. Its sum of digits was not less than k. But you were distracted a bit, and someone changed this number to n, replacing some digits with others. It's known that the length of the number didn't change.

You have to find the minimum number of digits in which these two numbers can differ.

Input

The first line contains integer k (1 ≤ k ≤ 10⁹).

The second line contains integer n (1 ≤ n < 10¹⁰⁰⁰⁰⁰).

There are no leading zeros in n. It's guaranteed that this situation is possible.

Output

Print the minimum number of digits in which the initial number and n can differ.

Examples

input

3
11

output

1

input

3
99

output

0

Note

In the first example, the initial number could be 12.

In the second example the sum of the digits of n is not less than k. The initial number could be equal to n.

解题思路：

由于题目过短题意就不说了，其实，具体思路看代码吧！

AC代码:

#include <stdio.h>
#include <stdlib.h>
#include <cmath>
#include <string.h>
#include <iostream>
#include<algorithm>
#include <queue>
#include <vector>
#include <cmath>
#include<string>
#define inf 0x3f3f3f3f
using namespace std;
const int maxn=1e5+10;
#define lrt (rt*2)
#define rrt  (rt*2+1)
#define LL long long
#define inf 0x3f3f3f3f
#define pi acos(-1.0)
#define exp 1e-8
#define For(i,n)  for(int i=0;i<n;i++)
//************************************************63
char a[maxn];
LL b[maxn];
int main()
{
    LL n,sum=0,k=0,ans=0;
    scanf("%I64d %s",&n,a);
    for(LL i=0;a[i]!='';i++)
    {
        b[k]=a[i]-'0';
        sum+=b[k];
        k++;
    }
    sort(b,b+k);
    while(sum<n)
    {
        sum+=(9-b[ans]);
        ans++;
    }
    cout<<ans<<endl;
    return 0;
}