Snackdown 2019 is coming! There are two rounds (round A and round B) after the qualification round. From both of them, teams can qualify to the pre-elimination round. According to the rules, in each of these two rounds, teams are sorted in descending order by their score and each team with a score greater or equal to the score of the team at the $K=1500$-th place advances to the pre-elimination round (this means it is possible to have more than $K$ qualified teams from each round in the case of one or more ties after the $K$-th place).
Today, the organizers ask you to count the number of teams which would qualify for the pre-elimination round from round A for a given value of $K$ (possibly different from $1500$). They provided the scores of all teams to you; you should ensure that all teams scoring at least as many points as the $K$-th team qualify.
-----Input-----
- The first line of the input contains a single integer $T$ denoting the number of test cases. The description of $T$ test cases follows.
- The first line of each test case contains two space-separated integers $N$ and $K$.
- The second line contains $N$ space-separated integers $S_1, S_2, \dots, S_N$.
-----Output-----
For each test case, print a single line containing one integer — the number of qualified teams.
-----Constraints-----
- $1 \le T \le 1,000$
- $1 \le K \le N \le 10^5$
- $1 \le S_i \le 10^9$ for each valid $i$
- the sum of $N$ for all test cases does not exceed $10^6$
-----Example Input-----
2
5 1
3 5 2 4 5
6 4
6 5 4 3 2 1
-----Example Output-----
2
4
t = int(input())
for i in range(t):
 n,k = map(int,input().split())
 l = list(map(int,input().split()))
 l.sort(reverse = True)
 a = l[k-1]
 j = 0
 count = 0
 while j<len(l):
  if l[j]>=a:
   count += 1
  else:
   break
  j += 1
 print(count)
t = int(input())
for i in range(t):
 n,k = map(int,input().split())
 l = list(map(int,input().split()))
 l.sort(reverse = True)
 a = l[k-1]
 j = 0
 count = 0
 while j<len(l):
  if l[j]>=a:
   count += 1
  else:
   break
  j += 1
 print(count)