158A. Next Round

*special problem/implementation, 800, http://codeforces.com/problemset/problem/158/A

"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 a~1~, a~2~, ..., 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.

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.

统计不小于第k位选手得分的人数

n, k = map(int, input().split())
score = [int(x) for x in input().split()]
num = 0
k_score = score[k-1]
for i in range(n):
	s = score[i]
	if s >= k_score and s>0:
		num += 1
print(num)

short code

i=lambda:map(int,input().split())
n,k=i()
a=list(i())
print(sum([x>=(a[k-1] or 1) for x in a]))

用or不能 用 and