1 1000000000 7
1000000000 9
 3 2 2
2 4
2 1
3 6
 2 1000000000 100000
1000000000 1000000
1000000000 10000
 1 100 100
50 70
 1 100 100
101 100

1
1000000000 8 2
2 3 1
1000000000 100001 1
99 100 1
101 100
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
 
using namespace __gnu_pbds;
using namespace std;

typedef long long int ll;
typedef unsigned long long int ull;
typedef long double ld;
typedef pair <ll, ll> pll;
typedef pair <int, int> pii;

typedef tree <ll, null_type, less <ll>, rb_tree_tag, tree_order_statistics_node_update> ordered_set;
// order_of_key(val): returns the number of values less than val
// find_by_order(k): returns an iterator to the kth largest element (0-based)

#define pb push_back
#define mp make_pair
#define ff first
#define ss second
#define all(a) a.begin(), a.end()
#define sz(a) (ll)(a.size())
#define endl "\n"

template <class Ch, class Tr, class Container>
basic_ostream <Ch, Tr> & operator << (basic_ostream <Ch, Tr> & os, Container const& x) 
{
    os << "{ ";
    for(auto& y : x) 
    {
        os << y << " ";
    }
    return os << "}";
}

template <class X, class Y>
ostream & operator << (ostream & os, pair <X, Y> const& p) 
{
    return os << "[" << p.ff << ", " << p.ss << "]";
}

ll gcd(ll a, ll b)
{
    if(b==0)
    {
        return a;
    }

    return gcd(b, a%b);
}

ll modexp(ll a, ll b, ll c)
{   
    a%=c;

    ll ans = 1;

    while(b)
    {
        if(b&1)
        {
            ans = (ans*a)%c;
        }

        a = (a*a)%c;
        b >>= 1;
    }

    return ans;
}

const ll L = 1e5+5;

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL); cout.tie(NULL);

    ll n, sx, sy;
    cin >> n >> sx >> sy;

    ll u = 0, d = 0, r = 0, l = 0;

    for(ll i=0; i<n; i++)
    {
        ll x, y;
        cin >> x >> y;
    
        if(y < sy)
        {
            d++;
        }

        else
        {
            u++;
        }

        if(x < sx)
        {
            l++;
        }

        else
        {
            r++;
        }
    }

    ll m = max(u, max(d, max(l, r)));

    cout << m << endl;

    if(m == u)
    {
        cout << sx << " " << sy+1 << endl;
    }

    else if(m == d)
    {
        cout << sx << " " << sy-1 << endl;
    }

    else if(m == r)
    {
        cout << sx+1 << " " << sy << endl;
    }

    else
    {
        cout << sx-1 << " " << sy << endl;
    }
    


    return 0;
}