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:
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"""


import sys
from functools import lru_cache, cmp_to_key
from heapq import merge, heapify, heappop, heappush, nsmallest
from math import ceil, floor, gcd, fabs, factorial, fmod, sqrt, inf
from collections import defaultdict as dd, deque, Counter as C
from itertools import combinations as comb, permutations as perm
from bisect import bisect_left as bl, bisect_right as br, bisect
from time import perf_counter
from fractions import Fraction
from decimal import Decimal
# sys.setrecursionlimit(pow(10, 6))
# sys.stdin = open("input.txt", "r")
# sys.stdout = open("output.txt", "w")
mod = pow(10, 9) + 7
# mod2 = 998244353
def data(): return sys.stdin.readline().strip()
def out(var): sys.stdout.write(str(var)+"\n")
def outa(*var, end="\n"): sys.stdout.write(' '.join(map(str, var)) + end)
def l(): return list(sp())
def sl(): return list(ssp())
def sp(): return map(int, data().split())
def ssp(): return map(str, data().split())
def l1d(n, val=0): return [val for i in range(n)]
def l2d(n, m, val=0): return [l1d(n, val) for j in range(m)]


n, m, i, j = sp()
i -= 1
j -= 1
vis = [[0] * m for i in range(n)]
other = set()
cnt = 0
temp = i
while temp < n:
    print(temp+1, j+1)
    other.add((temp, j))
    temp += 1
temp = i-1
while temp >= 0:
    print(temp+1, j+1)
    other.add((temp, j))
    temp -= 1
i = 0
if j != 0:
    j -= 1
    while j >= 0:
        print(i+1, j+1)
        other.add((i, j))
        j -= 1
j = 0
direction = 1
while cnt != n * m:
    if direction == 1:
        while j < m and not vis[i][j]:
            if (i, j) not in other:
                print(i+1, j+1)
            vis[i][j] = 1
            cnt += 1
            j += 1
        j -= 1
        i += 1
        direction = 2
    elif direction == 2:
        while i < n and not vis[i][j]:
            if (i, j) not in other:
                print(i+1, j+1)
            vis[i][j] = 1
            cnt += 1
            i += 1
        i -= 1
        j -= 1
        direction = 3
    elif direction == 3:
        while j >= 0 and not vis[i][j]:
            if (i, j) not in other:
                print(i+1, j+1)
            vis[i][j] = 1
            cnt += 1
            j -= 1
        j += 1
        i -= 1
        direction = 4
    else:
        while i >= 0 and not vis[i][j]:
            if (i, j) not in other:
                print(i+1, j+1)
            vis[i][j] = 1
            cnt += 1
            i -= 1
        i += 1
        j += 1
        direction = 1