A. Next Round

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"Contestant who earns a score equal to or greater than the k-th place finisher's score will advance to the next round, as long as the contestant earns a positive score..." — an excerpt from contest rules.

A total of n participants took part in the contest (n ≥ k), and you already know their scores. Calculate how many participants will advance to the next round.

Input

The first line of the input contains two integers n and k (1 ≤ k ≤ n ≤ 50) separated by a single space.

The second line contains n space-separated integers a₁, a₂, ..., a_n (0 ≤ a_i ≤ 100), where a_i is the score earned by the participant who got the i-th place. The given sequence is non-increasing (that is, for all i from 1 to n - 1 the following condition is fulfilled: a_i ≥ a_i + 1).

Output

Output the number of participants who advance to the next round.

   Sample test(s) 
 

     input 
   
复制代码8 5
10 9 8 7 7 7 5 5
8 5
10 9 8 7 7 7 5 5

     output 
   
复制代码6
6

     input 
   
复制代码4 2
0 0 0 0
4 2
0 0 0 0

     output 
   
复制代码0
0

Note

In the first example the participant on the 5th place earned 7 points. As the participant on the 6th place also earned 7 points, there are 6 advancers.

In the second example nobody got a positive score.

解题说明：水题，数组保存，循环搞定，注意0的情况

复制代码

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#include <iostream>
#include<cstdio>
using namespace std;

int main()
{
	int n,k;
	int a[51];
	int num=0;
	scanf("%d %d",&n,&k);
	for(int i=0;i<n;i++)
	{
		scanf("%d",&a[i]);
	}
	for(int i=0;i<n;i++)
	{
		if(a[i]>=a[k-1]&&a[i]!=0)
		{
			num++;
		}
	}

	printf("%dn",num);
	return 0;
}