codeforces158/A

Posted 2020-08-17 Pocker7

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 5
10 9 8 7 7 7 5 5

output

Copy

6

input

Copy

4 2
0 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.

 题意：长度为n的有序降序序列，成绩比第k名高或者相等的且不为0的人数

#include<iostream>
#include<string>
#include<cstring>
using namespace std;
typedef long long ll;
int a[1100];
int main()
{
    int n,k;
    cin>>n>>k;
    for(int i=1;i<=n;i++) cin>>a[i];
    int cnt=k;
    for(int i=k+1;i<=n;i++)
    {
        if(a[i]>=a[k]) cnt++; 
    }
    int tmp=cnt;
    for(int i=1;i<=tmp;i++) if(a[i]==0) cnt--;
    cout<<cnt;
    return 0;
 }