158A Next 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 and k (1 ≤ k ≤ n ≤ 50) separated by a single space.

The second line contains n space-separated integers a₁, a₂, ..., 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

Copy

8 510 9 8 7 7 7 5 5

Output

Copy

6

Input

Copy

4 20 0 0 0

Output

Copy

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.

提交次数：6次

错误原因：n>=k；数组中所有数必须是正的；第k位存入数组后-1；

 #include<iostream>
 using namespace std;
 int main()
 {
     ;
     cin>>n>>k;
     ];
     if(n>=k)
     {
         ;i<n;i++)
         {
             cin>>s[i];
         }
     }

         ;i<n;i++)
         {
             ]&&s[i]>)
             {
                 count++;
             }
         }
     cout<<count;

 }