#include <bits/stdc++.h>
#define ll long long
#define INF 1000000005
#define MOD 1000000007
#define EPS 1e-10
#define rep(i,n) for(int i=0;i<(int)n;++i)
#define each(a, b) for(auto (a): (b))
#define all(v) (v).begin(),(v).end()
#define fi first
#define se second
#define pb push_back
#define show(x) cout <<#x<<" = "<<(x)<<endl
#define spair(p) cout <<#p<<": "<<p.fi<<" "<<p.se<<endl
#define svec(v) cout<<#v<<":";rep(kbrni,v.size())cout<<" "<<v[kbrni];cout<<endl
#define sset(s) cout<<#s<<":";each(kbrni,s)cout <<" "<<kbrni;cout<<endl

using namespace std;

typedef pair<int,int>P;

const int MAX_N = 100005;

int main()
{
    int n;
    cin >> n;
    vector<int> pl,mn,ze;
    rep(i,2*n){
        int a;
        cin >> a;
        if(a > 0){
            pl.pb(a);
        }else if(a < 0){
            mn.pb(a);
        }else{
            ze.pb(a);
        }
    }
    sort(all(pl));
    sort(all(mn));
    int n1,n2;
    n1 = pl.size()-1;
    n2 = 0;
    int ans1 = 0;
    while(n1 >= 0 && n2 < mn.size()){
        if(pl[n1]+mn[n2] < 0){
            n1--;
            n2++;
            ans1++;
        }else{
            n1--;
        }
    }
    int cnt = (int)mn.size() - ans1;
    if(cnt <= (int)ze.size()){
        ans1 += cnt;
    }else{
        ans1 += (int)ze.size() + (cnt - (int)ze.size())/2;
    }
    n1 = 0;
    n2 = (int)mn.size()-1;
    int ans2 = 0;
    while(n1 < pl.size() && n2 >= 0){
        if(pl[n1] > -mn[n2]){
            n2--;
        }else if(pl[n1] < -mn[n2]){
            n1++;
        }else{
            ans2++;
            n1++;
            n2--;
        }
    }
    ans2 += (int)ze.size() / 2;
    n1 = (int)pl.size()-1;
    n2 = 0;
    int ans3 = 0;
    while(n1 >= 0 && n2 < (int)mn.size()){
        if(pl[n1]+mn[n2] > 0){
            n1--;
            n2++;
            ans3++;
        }else{
            n2++;
        }
    }
    cnt = (int)pl.size() - ans3;
    if(cnt <= (int)ze.size()){
        ans3 += cnt;
    }else{
        ans3 += (int)ze.size() + (cnt - (int)ze.size())/2;
    }
    cout << ans1 << " " << ans3 << " " << ans2 << endl;
    return 0;
}