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 io
import os

from collections import Counter, defaultdict, deque


def solveIntended(R, C, r0, c0):
    ans = []
    c = c0
    rs = list(range(R))
    rs.remove(r0)
    rs = [r0] + rs
    for r in rs:
        for i in range(C):
            ans.append(str(r + 1) + " " + str(c + 1))
            if i != C - 1:
                c = (c + 1) % C
    return "\n".join(ans)


def solve(R, C, r0, c0):
    # Solved wrong problem where rook takes one step at a time
    # Couldn't finish implementing this. It also has cases where it doesn't have solution
    # Fallback to intended solution for cases I didn't handle

    if R % 2 == 1 and C % 2 == 1:
        return solveIntended(R, C, r0, c0)

    def weave(minR, maxR, minC, maxC, r, c, direction):
        # Zigzag through a rectangle where the entry point is at one of the corners
        assert r in [minR, maxR]
        assert c in [minC, maxC]
        rows = list(range(minR, maxR + 1))
        cols = list(range(minC, maxC + 1))
        if r == maxR:
            rows = rows[::-1]
        if c == maxC:
            cols = cols[::-1]
        if direction == "h":
            for r in rows:
                for c in cols:
                    yield (r, c)
                cols = cols[::-1]
        else:
            assert direction == "v"
            for c in cols:
                for r in rows:
                    yield (r, c)
                rows = rows[::-1]

    def hamCycle(minR, maxR, minC, maxC):
        # If numRows or numCols is even, return a clockwise comb-shaped tour
        numRows = maxR - minR + 1
        numCols = maxC - minC + 1
        assert numRows >= 2 and numCols >= 2
        assert numRows % 2 == 0 or numCols % 2 == 0
        cycle = deque()
        if numRows % 2 == 0:
            # Go down first column
            (r, c) = (minR, minC)
            for r in range(minR, maxR + 1):
                cycle.appendleft((r, c))
            assert (r, c) == (maxR, minC)
            c += 1
            # Weave horizontally back to top
            assert (r, c) == (maxR, minC + 1)
            for r, c in weave(minR, maxR, minC + 1, maxC, r, c, "h"):
                cycle.appendleft((r, c))
            assert (r, c) == (minR, minC + 1)
        elif numCols % 2 == 0:
            # Go across first row
            (r, c) = (minR, minC)
            for c in range(minC, maxC + 1):
                cycle.append((r, c))
            assert (r, c) == (minR, maxC)
            r += 1
            # Weave vertically back to left
            assert (r, c) == (minR + 1, maxC)
            for r, c in weave(minR + 1, maxR, minC, maxC, r, c, "v"):
                cycle.append((r, c))
            assert (r, c) == (minR + 1, minC)
        else:
            assert False
        return cycle

    def reindex(cycle, r, c):
        # Rotate the deque so (r0, c0) is the first element
        index = cycle.index((r, c))
        cycle.rotate(-index)
        assert cycle[0] == (r, c)

    if R % 2 == 0 or C % 2 == 0:
        # Single cycle is enough
        ans = hamCycle(0, R - 1, 0, C - 1)
        reindex(ans, r0, c0)
    else:
        # R and C are both odd. Number of cells is odd.
        # If we color the chessboard, must start and end with same color. Not enough of the color if (r0 + c0) % 2 == 1
        assert (r0 + c0) % 2 == 0

        # Can probably split into 4 quadrants of:
        #   oo oe
        #   eo ee
        # Or
        #   ee eo
        #   oe oo
        # But this has too many edge cases!
        # TODO
        assert False

    return "\n".join(str(r + 1) + " " + str(c + 1) for r, c in ans)


if __name__ == "__main__":
    input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline

    N, M, X, Y = [int(x) for x in input().split()]
    X -= 1  # 0-indexed
    Y -= 1  # 0-indexed
    ans = solve(N, M, X, Y)
    print(ans)