Algorithm

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, ..., 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
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.

Code Examples

#1 Code Example with C++ Programming

```Code - C++ Programming```

``````#include<iostream>
using namespace std;

int main(){
int n;
int k;
int count;
cin>>n;
cin>>k;
int arr[n];
for(int i = 0; i < n; i++){
cout<<"i:"<<i<<" ";
cin>>arr[i];
}
int quali = 0, m;
if(arr[k-1] == 0){

for(m = k-2; m >= 0; m--){

if(arr[m] > 0){

quali = arr[m];
count  = m+1;
cout<<count;
return 0;
}
}
if(quali == 0){
cout<<0;
return 0;
}
}
else{
quali = arr[k-1];
count = k;
}
cout<<endl;
int j = count-2;
while(true){
if(arr[j] == quali){
count++;
}
else{
break;
}
j--;
}
cout<<count;
return 0;
}``````
Copy The Code &

Input

cmd
8 5
10 9 8 7 7 7 5 5

Output

cmd
6