Next Round Solution || Codeforces

Java solution for codeforces problem 158A. The problem is "Next Round" on codeforces (a competitive coding platform).

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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

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.

Solution : in Java

import java.util.Scanner;

public class NextRound {

    public static void main(String[] args) {

        Scanner sc = new Scanner(System.in);

        int n = sc.nextInt();

        int k = sc.nextInt();

        int arr[] = new int[n];

        

        for(int i=0;i<n;i++){

            arr[i] = sc.nextInt();

        }

        

        if(arr[k-1]>0){

            while(k<n){

                if(arr[k-1]>arr[k]){

                    break;

                }

                k++;

            }

        }

        else{

            while(k>0){

                if(arr[k-1]>0){

                    break;

                }

                k--;

            }

        }

        

        System.out.println(k);

    }

}

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