Next Round

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我是靠谱客的博主知性豆芽，最近开发中收集的这篇文章主要介绍Next Round，觉得挺不错的，现在分享给大家，希望可以做个参考。

概述

"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, ..., an (0 ≤ ai ≤ 100), where ai 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: ai ≥ ai + 1).

Output

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

Examples

input
8 5
10 9 8 7 7 7 5 5

output
6

input
4 2
0 0 0 0

output
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.

ACM的一道水题，就是统计大于等于第k且不等于零的数的个数。

</pre><pre name="code" class="cpp">#include "stdio.h"
#define N 51
int main()
{
int n,k,i,count=0;
scanf ("%d%d",&n,&k);
int a[N];
for (i=0;i<n;i++)
scanf ("%d",&a[i]);
for (i=0;i<n;i++)
{
if (a[i]&&a[i]>=a[k-1])
count++;
}
printf ("%dn",count);
}