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 JAVA):

Solution:
import java.lang.reflect.Array;
import java.util.Arrays;
import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;

public class Main {
    /*code forces 1395/B Boboniu Plays Chess
        在一个n*m的地图内
     */
    public static void main(String[] args) {
        Scanner scr = new Scanner(System.in);
        int rows = scr.nextInt();
        int columns = scr.nextInt();
        int Sx = scr.nextInt() - 1;
        int Sy = scr.nextInt() - 1;
        int[] position = new int[]{Sx,Sy};
        while (position[0] >= 0) {
            print(position);
            position = move(position, Position.LEFT);
        }
        position[0] = Sx + 1;
        while (position[0] < rows) {
            print(position);
            position = move(position, Position.RIGHT);
        }
        position = move(position, Position.LEFT);
        position[1] = Sy-1;
        boolean inLeft = false;
        Position p = Position.UP;
        while (position[1] < columns) {
            if (inLeft) {
                while (position[0] < rows) {
                    print(position);
                    position = move(position, Position.RIGHT);
                }
                position = move(position, Position.LEFT);
            } else {
                while (position[0] >= 0) {
                    print(position);
                    position = move(position, Position.LEFT);
                }
                position = move(position, Position.RIGHT);
            }
            inLeft = !inLeft;
            if (position[1] == 0) {
                position[1] = Sy;
                p = Position.DOWN;
            }
            position = move(position, p);
        }

    }
    static int[] move(int[] ps,Position p) {
        return move(ps[0],ps[1],p);
    }
    static int[] move(int x, int y,Position p) {
        switch (p){
            case UP:
                return new int[]{x,y-1};
            case DOWN:
                return new int[]{x,y+1};
            case LEFT:
                return new int[]{x-1,y};
            case RIGHT:
                return new int[]{x+1,y};
        }
        return new int[]{x,y};
    }

    public enum Position {
        UP, DOWN, LEFT, RIGHT
    }

    static void print(int[] position) {
        print(position[0],position[1]);
    }

    static void print(int x, int y) {
        System.out.println((x+1) + " " + (y+1));
    }

    /*code forces 1399/C Boats Competition
            给出一批数字，任意两个数字的值相加组成一个组，问最多能有几个组同时存在（即最多有几个组能值相等）
            测试用例：
            1 2 2 1 1 2 //3
            6 6 6 6 6 6 6 8 //3
            1 //0
     */
//    public static void main(String[] args) {
//        Scanner scr = new Scanner(System.in);
//        int n = scr.nextInt();
//        for (int i = 0; i < n; i++) {
//            int numCount = Integer.valueOf(scr.next());
//            int[] nums = new int[numCount];
//            for (int j = 0; j < numCount; j++) {
//                nums[j] = scr.nextInt();
//            }
//
//            System.out.println(getMaxCount(nums));
//        }
//    }
//
//    static int getMaxCount(int[] nums) {
//        int maxResult = 0;
//        int min = nums[0], max = nums[0];
//        Map<Integer, Integer> map = new HashMap<>();
//        for (int thisN : nums) {
//            min = thisN < min ? thisN : min;
//            max = thisN > max ? thisN : max;
//            map.put(thisN, map.getOrDefault(thisN, 0) + 1);
//        }
//        Integer[] keys =  map.keySet().toArray(new Integer[0]);
//        Arrays.sort(keys);
//        for (int sum = 2 * min; sum <= 2 * max; sum++) {
//            int thisResult = 0;
//            for (int key : keys) {
//                if (sum < 2*key) {
//                    break;
//                } else if (2 * key == sum) {
//                    thisResult += map.get(key) / 2;
//                    break;
//                } else if (map.containsKey(sum - key)) {
//                    thisResult += Math.min(map.get(key), map.get(sum - key));
//                }
//
//            }
//            maxResult = Math.max(thisResult, maxResult);
//        }
//        return maxResult;
//    }

    /*code forces 71/A Way Too Long Words
        一个单词超过10个字符的话，则将中间的字符用数字展示，如：localization->l10n；internationalization->i18n
     */
//    public static void main(String[] args) {
//        Scanner scr = new Scanner(System.in);
//        int n = scr.nextInt();
//        for (int i=0;i<n;i++){
//            String s = scr.next();
//            System.out.println(s.length()>10?s.substring(0,1)+(s.length()-2)+s.substring(s.length()-1):s);
//        }
//    }

//    /*code forces /1/A Theatre Square
//        一个广场，长宽分别为m和n，用长为a的正方体石板铺满，可以超出广场面积，石板不能破坏
//     */
//    public static void main(String[] args) {
//        Scanner scr = new Scanner(System.in);
//        double m = scr.nextDouble();
//        double n = scr.nextDouble();
//        double a = scr.nextDouble();
//        double w = m/a;
//        double l = n/a;
//        System.out.print(Math.ceil(w)*Math.ceil(l));
//    }

    /*code forces

     */
//    public static void main(String[] args) {
//        Scanner scr = new Scanner(System.in);
//        int weight = scr.nextInt();
//        System.out.print(weight%2==0&&weight>2?"YES":"NO");
//    }

}