human
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.
#
assistant
#Here is code to 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.

# Maximum Size Square Sub-Matrix with all 1s |
#
# Given a binary matrix, find out the maximum
# size square sub-matrix with all 1s.
#
# For example, consider the below binary matrix.
#
# """
#
#
#
# def printMaxSubSquare(M):
#
#     R = len(M) #no of rows
#     C = len(M[0]) #no of columns
#
#     S = [[0 for k in range(C)] for l in range(R)]
#     # here we have set the first row and columns of
#     # s[][]
#
#     # Construct other entries
#     for i in range(1,R):
#         for j in range(1,C):
#             if M[i][j]==1:
#                 S[i][j]= min(S[i-1][j],S[i][j-1],S[i-1][j-1])+1
#             else:
#                 S[i][j] = 0
#
#     max = -1
#     for i in S:
#         for j in i:
#             if j>max:
#                 max = j
#     return max
# M = [[0, 1, 1, 0, 1],
#     [1, 1, 0, 1, 0],
#     [0, 1, 1, 1, 0],
#     [1, 1, 1, 1, 0],
#     [1, 1, 1, 1, 1],
#     [0, 0, 0, 0, 0]]
#
# print(printMaxSubSquare(M))
n,k = map(int,input().split())
arr = list(map(int,input().split()))
min = arr[k-1]
count = 0
for i in arr:
    if i <=0:
        count+=0
    elif i >= min:
        count+=1
print (count)