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.util.*;
import java.io.*;

public class Main {
	static final int M = 1000000007;

	static class FastScanner {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer("");

		String next() {
			while (!st.hasMoreTokens())
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();
		}

		int nextInt() {
			return Integer.parseInt(next());
		}

		long[] readArrayLong(int n) {
			long[] a = new long[n];
			for (int i = 0; i < n; i++)
				a[i] = nextLong();
			return a;
		}

		int[] readArrayInt(int n) {
			int[] a = new int[n];
			for (int i = 0; i < n; i++)
				a[i] = nextInt();
			return a;
		}

		long nextLong() {
			return Long.parseLong(next());
		}
	}

	// sorting template
	static void sortLong(long[] a) {
		ArrayList<Long> l = new ArrayList<>();
		for (long i : a)
			l.add(i);
		Collections.sort(l);
		for (int i = 0; i < a.length; i++)
			a[i] = l.get(i);
	}

	static void sortInt(int[] a) {
		ArrayList<Integer> l = new ArrayList<>();
		for (int i : a)
			l.add(i);
		Collections.sort(l);
		for (int i = 0; i < a.length; i++)
			a[i] = l.get(i);
	}

	static class Graph {
		private ArrayList<Integer> adj[];
		public String s;
		public String dpch;

		Graph(int n) {
			adj = new ArrayList[n + 1];
			for (int i = 1; i <= n; i++) {
				adj[i] = new ArrayList<>();
			}
		}

		void addEdges(int u, int v) {
			adj[u].add(v);
			adj[v].add(u);
		}

		void dfs(int v, boolean[] vis, int d) {
			vis[v] = true;
			if (d > maxDist) {
				maxNode = v;
				maxDist = d;
			}
			for (int u : adj[v]) {
				if (!vis[u]) {
					dfs(u, vis, d + 1);
				}
			}
		}

		void bfs(int v, boolean[] vis) {
			Queue<Integer> q = new LinkedList<>();
			q.add(v);
			dist[v] = 0;
			vis[v] = true;
			while (q.size() > 0) {
				int node = q.poll();
				for (int u : adj[node]) {
					if (!vis[u]) {
						vis[u] = true;
						dist[u] = dist[node] + 1;
						q.add(u);
					}
				}
			}
		}
	}

	static class Pair<F, S> {
		F first;
		S second;

		Pair(F first, S second) {
			this.first = first;
			this.second = second;
		}

		public F getFirst() {
			return this.first;
		}

		public S getSecond() {
			return this.second;
		}
		// Sort on basis of first term, reverse p1.first and p2.first to sort in desc
		// Collections.sort(a, (p1, p2) -> (p1.first - p2.first));

		// Sort on the basis of second term, reverse p1.second and p2.second to sort in
		// desc
		// Collections.sort(a, (p1, p2) -> (p1.second - p2.second));
	}

	static int __gcd(int a, int b) {
		if (b == 0)
			return a;
		return __gcd(b, a % b);
	}

	public static int subtree, maxNode, maxDist;
	public static int[] dist;
	public static int[] connected;

	public static void main(String[] args) throws IOException {
		FastScanner fs = new FastScanner();
		int t = 1;
		StringBuffer res = new StringBuffer();
		for (int tt = 1; tt <= t; tt++) {
			int n = fs.nextInt();
			int m = fs.nextInt();
			int x = fs.nextInt();
			int y = fs.nextInt();
			boolean[][] vis = new boolean[n + 1][m + 1];
			res.append(x + " " + y).append("\n");
			vis[x][y] = true;
			res.append("1" + " " + y).append("\n");
			vis[1][y] = true;
			for (int i = 1; i <= n; i++) {
				if (i % 2 != 0) {
					for (int j = 1; j <= m; j++) {
						if (!vis[i][j]) {
							res.append(i + " " + j).append("\n");
							vis[i][j] = true;
						}
					}
				} else {
					for (int j = m; j >= 1; j--) {
						if (!vis[i][j]) {
							res.append(i + " " + j).append("\n");
							vis[i][j] = true;
						}
					}
				}
			}
		}
		System.out.println(res);
	}
}