1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
#include <bits/stdc++.h>
//#pragma GCC optimize("Ofast")
//#pragma GCC optimize ("unroll-loops")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")

//szablon czesciowo skopiowany z profilu https://codeforces.com/profile/Geothermal
using namespace std;
 
typedef long long LL;
typedef long double LD;
 
typedef pair<int, int> pii;
typedef pair<LL,LL> pll;
typedef pair<LD,LD> pdd;
 
typedef vector<int> vi;
typedef vector<bool> vb;
typedef vector<LD> vld;
typedef vector<LL> vll;
typedef vector<pii> vpii;
typedef vector<pll> vpll;
 
template<class T> using pq = priority_queue<T>;
template<class T> using pqg = priority_queue<T, vector<T>, greater<T>>;
 
#define rep(i, a, b) for (int i=a; i<(b); i++)
#define repd(i,a,b) for (int i = (a); i >= b; i--)
#define sz(x) (int)(x).size()
#define pb push_back
#define st first
#define nd second
#define lb lower_bound
#define ub upper_bound
#define all(x) x.begin(), x.end()
#define memo(x) memset(x, 0, sizeof(x))
#define debug(x) cerr << x << " "
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
LL losuj(LL a, LL b){ return a+rng()%(b-a+1);}

void solve(){
	int n, k; cin >> n >> k;
	vi t(n);
	rep(i,0,n) cin >> t[i];
	sort(all(t), greater<int>());
	while(k<n && t[k]==t[k-1]) k++;
	cout << k << "\n";
}

int main(){
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	
	int t = 1;
	//cin >> t;
	while (t--) solve();
	
	return 0;
}