CF 158A Round

A. Next Round

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

"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 andk (1 ≤ k ≤ n ≤ 50) separated by a single space.

The second line contains n space-separated integersa₁, a₂, ..., a_n (0 ≤ a_i ≤ 100), wherea_i is the score earned by the participant who got thei-th place. The given sequence is non-increasing (that is, for alli 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.

Examples

Input

8 510 9 8 7 7 7 5 5

Output

6

Input

4 20 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.

题意：有一组数据，输入两个数字n，k；n代表数字个数，如果这些数据大于等于第K个数字，则通过人数加一

PS：CF如果没有要求多实例，就单实例，否则会WA；坑我好久；

#include <bits/stdc++.h>using namespace std ;int dp[101];int main(){int n , k ;cin>>n>>k; for(int i = 1 ; i <= n ; i++){cin>>dp[i];}int sum=0;for(int i = 1 ; i <=n;i++){if(dp[i]>=dp[k]&&dp[i]!=0){sum++;}}cout<<sum<<endl;return 0 ;}