Polycarp is crazy about round numbers. He especially likes the numbers divisible by 10k.In the given number of n Polycarp wants to remove the least number of digits to get a number that is divisible by 10k. For example, if k?=?3, in the number 30020 it is enough to delete a single digit (2). In this case, the result is 3000 that is divisible by 103?=?1000.Write a program that prints the minimum number of digits to be deleted from the given integer number n, so that the result is divisible by 10k. The result should not start with the unnecessary leading zero (i.e., zero can start only the number 0, which is required to be written as exactly one digit).It is guaranteed that the answer exists.

Input

The only line of the input contains two integer numbers n and k (0?≤?n?≤?2?000?000?000, 1?≤?k?≤?9).It is guaranteed that the answer exists. All numbers in the input are written in traditional notation of integers, that is, without any extra leading zeros.

Output

Print w — the required minimal number of digits to erase. After removing the appropriate w digits from the number n, the result should have a value that is divisible by 10k. The result can start with digit 0 in the single case (the result is zero and written by exactly the only digit 0).

Example

Input
30020 3
Output
1
Input
100 9
Output
2
Input
10203049 2
Output
3

題目大意

給一個數(shù)字n，從中移除一些位上的數(shù)，使其能夠被10^k整除，要求移除次數(shù)最小，題目保證有解

解題思路

若是n中0的個數(shù)小于k，那么就需要把其他數(shù)字都移除，只剩下一個0，否則就從后往前掃描，若不是0，++ans,移除，是0，設(shè)置一個計數(shù)器i來統(tǒng)計0的個數(shù)，當(dāng)0的個數(shù)達(dá)到k時，此時就剛好可以被10^k整除，輸出ans

代碼

#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; typedef long long LL; const int Maxn = 1e3;int main() {int ans = 0;char a[11];memset(a, 0, sizeof a);int k;scanf("%s%d",a,&k);int len = strlen(a);int num = 0;for(int i = 0; i < len; ++i)if(a[i] == '0')++num;if(num < k)ans = len - 1;else{int i = 0, j = len - 1;while(i < k){if(a[j] == '0')++i;else++ans;--j; } }printf("%d\n",ans);return 0; }

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