#pragma GCC optimization ("O3")

#include <bits/stdc++.h>

using namespace std;

using ll = long long;
using vec = vector<ll>;
using mat = vector<vec>;
using pll = pair<ll,ll>;

#define INF (1LL<<61)
//#define MOD 1000000007LL 
#define MOD 998244353LL
#define EPS (1e-10)

#define PR(x) cout << (x) << endl
#define PS(x) cout << (x) << " "
#define REP(i,m,n) for(ll (i)=(m),(i_len)=(n);(i)<(i_len);++(i))
#define FORE(i,v) for(auto (i):v)
#define ALL(x) (x).begin(), (x).end()
#define SZ(x) ((ll)(x).size())
#define REV(x) reverse(ALL((x)))
#define ASC(x) sort(ALL((x)))
#define DESC(x) {ASC((x)); REV((x));}
#define BIT(s,i) (((s)>>(i))&1)
#define pb push_back
#define fi first
#define se second

template<class T> inline int chmin(T& a, T b) {if(a>b) {a=b; return 1;} return 0;}
template<class T> inline int chmax(T& a, T b) {if(a<b) {a=b; return 1;} return 0;}
class mint {
public:
    ll x;
    mint(ll x=0) : x((x%MOD+MOD)%MOD) {}
    mint operator-() const {return mint(-x);}
    mint& operator+=(const mint& a) {if((x+=a.x)>=MOD) x-=MOD; return *this;}
    mint& operator-=(const mint& a) {if((x+=MOD-a.x)>=MOD) x-=MOD; return *this;}
    mint& operator*=(const mint& a) {(x*=a.x)%=MOD; return *this;}
    mint operator+(const mint& a) const {mint b(*this); return b+=a;}
    mint operator-(const mint& a) const {mint b(*this); return b-=a;}
    mint operator*(const mint& a) const {mint b(*this); return b*=a;}
    mint pow(ll t) const {if(!t) return 1; mint a=pow(t>>1); return (t&1?*this*a:a)*a;}
    mint inv() const {return pow(MOD-2);}
    mint& operator/=(const mint& a) {return *this*=a.inv();}
    mint operator/(const mint& a) const {mint b(*this); return b/=a;}
};
istream &operator>>(istream& is, mint& a) {ll t; is>>t; a=t; return is;}
ostream &operator<<(ostream& os, const mint& a) {return os<<a.x;}

template <typename T> struct BinaryIndexedTree {

    int n;
    vector<T> dat;
    
    BinaryIndexedTree(int n_) {
        n = n_+1;
        dat = vector<T>(n, 0);
    }

    void add(int i, T x) {
        for(int idx=i; idx<n; idx+=(idx&-idx)) dat[idx] += x;
    }

    T sum(int i) {
        T s(0);
        for(int idx=i; idx>0; idx-=(idx&-idx)) s += dat[idx];
        return s;
    }

    int lower_bound(T w) {
        if(w <= 0) return 0;
        else {
            int x = 0, r = 1;
            while(r < n) r <<= 1;
            for(int len=r; len>0; len>>=1) {
                if(x+len < n && dat[x+len] < w) {
                    w -= dat[x+len];
                    x += len;
                }
            }
            return x+1;
        }
    }

};

int main()
{
    ll N, K;
    cin >> N >> K;

    ll M = 1000000;
    BinaryIndexedTree<ll> bit(M+1);
    REP(i,0,N) {
        ll w;
        cin >> w;
        if(w > 0) {
            if(bit.sum(M+1)-bit.sum(w-1) < K) bit.add(w, 1);
        }
        else {
            w = -w;
            if(bit.sum(w)-bit.sum(w-1) >= 1) bit.add(w, -1);
        }
    }
    PR(bit.sum(M+1));

    return 0;
}

/*



*/