Problem: Boboniu likes playing chess with his employees. As we know, no employee can beat the boss in the chess game, so Boboniu has never lost in any round.

You are a new applicant for his company. Boboniu will test you with the following chess question:

Consider a n× m grid (rows are numbered from 1 to n, and columns are numbered from 1 to m). You have a chess piece, and it stands at some cell (S_x,S_y) which is not on the border (i.e. 2 ≤ S_x ≤ n-1 and 2 ≤ S_y ≤ m-1).

From the cell (x,y), you can move your chess piece to (x,y') (1≤ y'≤ m, y' ≠ y) or (x',y) (1≤ x'≤ n, x'≠ x). In other words, the chess piece moves as a rook. From the cell, you can move to any cell on the same row or column.

Your goal is to visit each cell exactly once. Can you find a solution?

Note that cells on the path between two adjacent cells in your route are not counted as visited, and it is not required to return to the starting point.

Input

The only line of the input contains four integers n, m, S_x and S_y (3≤ n,m≤ 100, 2 ≤ S_x ≤ n-1, 2 ≤ S_y ≤ m-1) — the number of rows, the number of columns, and the initial position of your chess piece, respectively.

Output

You should print n⋅ m lines.

The i-th line should contain two integers x_i and y_i (1 ≤ x_i ≤ n, 1 ≤ y_i ≤ m), denoting the i-th cell that you visited. You should print exactly nm pairs (x_i, y_i), they should cover all possible pairs (x_i, y_i), such that 1 ≤ x_i ≤ n, 1 ≤ y_i ≤ m.

We can show that under these constraints there always exists a solution. If there are multiple answers, print any.

Examples

Input


3 3 2 2


Output


2 2
1 2
1 3
2 3
3 3
3 2
3 1
2 1
1 1


Input


3 4 2 2


Output


2 2
2 1
2 3
2 4
1 4
3 4
3 3
3 2
3 1
1 1
1 2
1 3

Note

Possible routes for two examples:

<image>

Solution (in PYTHON3):

Solution:
import math, sys
from collections import defaultdict, Counter, deque
 
INF = float('inf')
MOD = 1000000007
 
def gcd(a, b):
	while b:
		a, b = b, a%b
	return a
 
def isPrime(n):
	if (n <= 1): 
		return False
	i = 2
	while i ** 2 <= n:
		if n % i == 0:
			return False
		i += 1
	return True
 
def primeFactors(n):
	factors = []
	i = 2
	while i ** 2 <= n:
		while n % i == 0:
			factors.append(i)
			n //= i 
		i += 1
	if n > 1:
		factors.append(n)
	return factors
 
def vars():
	return map(int, input().split())
 
def array():
	return list(map(int, input().split()))


def a():
	r, g, b, w = vars()
	colors = [r, g, b, w]

	odd = 0
	for color in colors:
		if color % 2 == 1:
			odd += 1

	if (r == 0 or g == 0 or b == 0) and odd > 1:
		print('No')
		return

	if colors[-1] % 2 == 1:
		odd -= 1 

	if odd == 3:
		print('Yes')
	elif odd == 2 and colors[-1] % 2 == 1:
		print('Yes')
	elif odd == 0 or (odd == 1 and colors[-1] % 2 == 0):
		print('Yes')
	else:
		print('No')

def b():
	n, m, start, end = vars()

	visited = {}
	
	x, y = start, end

	while x >= 1:
		print(x, y)
		x -= 1 

	x += 1 
	side = 0
	for x in range(1, n + 1):
		if not side:
			r = (1, m)
		else:
			r = (m, 1, -1)

		for y in range(*r):
			if (x <= start and y != end) or x > start:
				print(x, y)

		print(x, r[1])
		visited[(x, r[1])] = True
		side ^= 1

def most_common_set_bits(n1, n2):
	cnt = 0
	i = min(len(n1), len(n2)) - 1
	while i >= 0:
		if n1[i] == n2[i]:
			cnt += 1 
		i -= 1
	return cnt


if __name__ == "__main__":
	t = 1
	# t = int(input())
	for _ in range(t):
		b()