The map of the capital of Berland can be viewed on the infinite coordinate plane. Each point with integer coordinates contains a building, and there are streets connecting every building to four neighbouring buildings. All streets are parallel to the coordinate axes.

The main school of the capital is located in (s_x, s_y). There are n students attending this school, the i-th of them lives in the house located in (x_i, y_i). It is possible that some students live in the same house, but no student lives in (s_x, s_y).

After classes end, each student walks from the school to his house along one of the shortest paths. So the distance the i-th student goes from the school to his house is |s_x - x_i| + |s_y - y_i|.

The Provision Department of Berland has decided to open a shawarma tent somewhere in the capital (at some point with integer coordinates). It is considered that the i-th student will buy a shawarma if at least one of the shortest paths from the school to the i-th student's house goes through the point where the shawarma tent is located. It is forbidden to place the shawarma tent at the point where the school is located, but the coordinates of the shawarma tent may coincide with the coordinates of the house of some student (or even multiple students).

You want to find the maximum possible number of students buying shawarma and the optimal location for the tent itself.

Input

The first line contains three integers n, s_x, s_y (1 ≤ n ≤ 200 000, 0 ≤ s_x, s_y ≤ 10^{9}) — the number of students and the coordinates of the school, respectively.

Then n lines follow. The i-th of them contains two integers x_i, y_i (0 ≤ x_i, y_i ≤ 10^{9}) — the location of the house where the i-th student lives. Some locations of houses may coincide, but no student lives in the same location where the school is situated.

Output

The output should consist of two lines. The first of them should contain one integer c — the maximum number of students that will buy shawarmas at the tent. 

The second line should contain two integers p_x and p_y — the coordinates where the tent should be located. If there are multiple answers, print any of them. Note that each of p_x and p_y should be not less than 0 and not greater than 10^{9}.

Examples

Input


4 3 2
1 3
4 2
5 1
4 1


Output


3
4 2


Input


3 100 100
0 0
0 0
100 200


Output


2
99 100


Input


7 10 12
5 6
20 23
15 4
16 5
4 54
12 1
4 15


Output


4
10 11

Note

In the first example, If we build the shawarma tent in (4, 2), then the students living in (4, 2), (4, 1) and (5, 1) will visit it.

In the second example, it is possible to build the shawarma tent in (1, 1), then both students living in (0, 0) will visit it.
import sys
import os
from io import BytesIO, IOBase

#########################
# imgur.com/Pkt7iIf.png #
#########################

# returns the list of prime numbers less than or equal to n:
'''def sieve(n):
    if n < 2: return list()
    prime = [True for _ in range(n + 1)]
    p = 3
    while p * p <= n:
        if prime[p]:
            for i in range(p * 2, n + 1, p):
                prime[i] = False
        p += 2
    r = [2]
    for p in range(3, n + 1, 2):
        if prime[p]:
            r.append(p)
    return r'''

# returns all the divisors of a number n(takes an additional parameter start):
'''def divs(n, start=1):
    divisors = []
    for i in range(start, int(math.sqrt(n) + 1)):
        if n % i == 0:
            if n / i == i:
                divisors.append(i)
            else:
                divisors.extend([i, n // i])
    return divisors'''

# returns the number of factors of a given number if a primes list is given:
'''def divn(n, primes):
    divs_number = 1
    for i in primes:
        if n == 1:
            return divs_number
        t = 1
        while n % i == 0:
            t += 1
            n //= i
        divs_number *= t
    return divs_number'''

# returns the leftmost and rightmost positions of x in a given list d(if x isnot present then returns (-1,-1)):
'''def flin(d, x, default=-1):
    left = right = -1
    for i in range(len(d)):
        if d[i] == x:
            if left == -1: left = i
            right = i
    if left == -1:
        return (default, default)
    else:
        return (left, right)'''

# returns (b**p)%m 
'''def modpow(b,p,m):
    res=1
    b=b%m
    while(p):
        if p&1:
            res=(res*b)%m
        b=(b*b)%m
        p>>=1
    return res'''

# if m is a prime this can be used to determine (1//a)%m or modular multiplicative inverse of a number a
'''def mod_inv(a,m):
    return modpow(a,m-2,m)'''

# returns the ncr%m for (if m is a prime number) for very large n and r
'''def ncr(n,r,m):
    res=1
    if r==0:
        return 1
    if n-r<r:
        r=n-r
    p,k=1,1
    while(r):
        res=((res%m)*(((n%m)*mod_inv(r,m))%m))%m
        n-=1
        r-=1
    return res'''

# returns ncr%m (if m is a prime number and there should be a list fact which stores the factorial values upto n):
'''def ncrlis(n,r,m):
    return (fact[n]*(mod_inv(fact[r],m)*mod_inv(fact[n-r],m))%m)%m'''

#factorial list which stores factorial of values upto n:
'''mx_lmt=10**5+10
fact=[1 for i in range(mx_lmt)]
for i in range(1,mx_lmt):
    fact[i]=(i*fact[i-1])%mod'''

#count xor of numbers from 1 to n:
'''def xor1_n(n):
    d={0:n,1:1,2:n+1,3:0}
    return d[n&3]'''

def cel(n, k): return n // k + (n % k != 0)
def ii(): return int(input())
def mi(): return map(int, input().split())
def li(): return list(map(int, input().split()))
def lcm(a, b): return abs(a * b) // math.gcd(a, b)
def prr(a, sep=' '): print(sep.join(map(str, a)))
def dd(): return defaultdict(int)
def ddl(): return defaultdict(list)
def ddd(): return defaultdict(defaultdict(int))
 

# region fastio
 
BUFSIZE = 8192
 
 
class FastIO(IOBase):
	newlines = 0
 
	def __init__(self, file):
		self._fd = file.fileno()
		self.buffer = BytesIO()
		self.writable = "x" in file.mode or "r" not in file.mode
		self.write = self.buffer.write if self.writable else None
 
	def read(self):
			while True:
					b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
					if not b:
							break
					ptr = self.buffer.tell()
					self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
			self.newlines = 0
			return self.buffer.read()
 
	def readline(self):
			while self.newlines == 0:
					b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
					self.newlines = b.count(b"\n") + (not b)
					ptr = self.buffer.tell()
					self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
			self.newlines -= 1
			return self.buffer.readline()
 
	def flush(self):
			if self.writable:
					os.write(self._fd, self.buffer.getvalue())
					self.buffer.truncate(0), self.buffer.seek(0)
 
 
class IOWrapper(IOBase):
	def __init__(self, file):
			self.buffer = FastIO(file)
			self.flush = self.buffer.flush
			self.writable = self.buffer.writable
			self.write = lambda s: self.buffer.write(s.encode("ascii"))
			self.read = lambda: self.buffer.read().decode("ascii")
			self.readline = lambda: self.buffer.readline().decode("ascii")
 

sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")

from collections import defaultdict 
#from collections import deque
#from collections import OrderedDict
#from math import gcd
#import time
#import itertools
#import timeit
#import random
#from bisect import bisect_left as bl
#from bisect import bisect_right as br
#from bisect import insort_left as il
#from bisect import insort_right as ir
#from heapq import *
#mod=998244353
#mod=10**9+7

for _ in range(1):
    n,sx,sy=mi()
    r,l,u,d=0,0,0,0
    for i in range(n):
        x,y=mi()
        if x>sx:
            r+=1
        elif x<sx:
            l+=1
        if y>sy:
            u+=1
        elif y<sy:
            d+=1
    if r==max(l,r,u,d):
        print(r)
        print(sx+1,sy)
    elif l==max(l,r,u,d):
        print(l)
        print(sx-1,sy)
    elif u==max(l,r,u,d):
        print(u)
        print(sx,sy+1)
    else:
        print(d)
        print(sx,sy-1)