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

public class A {

	public static void main(String[] args) throws Exception {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int m = sc.nextInt();
		int x = sc.nextInt();
		int y = sc.nextInt();
		boolean vis[][] = new boolean[n][m];

		for (int i = y - 1; i >= 0; i--) {
			System.out.println(x + " " + (i + 1));
			vis[x-1][i] = true;
		}
		for (int i = 0; i < m; i++) {
			if (i % 2 == 0)
				for (int j = 0; j < n; j++) {
					if (!vis[j][i]) {
						System.out.println((j + 1) + " " + (i + 1));
						vis[j][i] = true;
					}
				}
			else {
				for (int j = n-1; j >= 0; j--) {
					if (!vis[j][i]) {
						System.out.println((j + 1) + " " + (i + 1));
						vis[j][i] = true;
					}
				}
			}
		}
	}

	/////////////////////////////////////////////////////////////////////////////////////////////

	static class pair implements Comparable<pair> {
		int x;
		long y;
		int c;

		pair(int s, long d, int e) {
			x = s;
			y = (int) d;
			c = e;
		}

		@Override
		public int compareTo(pair p) {
			return Long.compare(p.y, y);
		}

		@Override
		public String toString() {
			return x + " " + y;

		}
	}

	static long mod(long ans, int mod) {
		return (ans % mod + mod) % mod;
	}

	static long gcd(long a, long b) {
		if (a == 0)
			return b;
		return gcd(b % a, a);
	}

	static long lcm(long a, long b) {
		return (a * b) / gcd(a, b);
	}

	public static int log(int n, int base) {
		int ans = 0;
		while (n + 1 > base) {
			ans++;
			n /= base;
		}
		return ans;
	}

	static long pow(long b, long e) {
		long ans = 1;
		while (e > 0) {
			if ((e & 1) == 1)
				ans = ((ans * 1l * b));
			e >>= 1;
			{

			}
			b = ((b * 1l * b));
		}
		return ans;
	}

	static int powmod(int b, long e, int mod) {
		int ans = 1;
		b %= mod;
		while (e > 0) {
			if ((e & 1) == 1)
				ans = (int) ((ans * 1l * b) % mod);
			e >>= 1;
			b = (int) ((b * 1l * b) % mod);
		}
		return ans;
	}

	static int ceil(int a, int b) {
		int ans = a / b;
		return a % b == 0 ? ans : ans + 1;
	}

	static long ceil(long a, long b) {
		long ans = a / b;
		return a % b == 0 ? ans : ans + 1;
	}

	// Returns nCr % p
	static int nCrModp(int n, int r, int p) {
		if (r > n - r)
			r = n - r;

		// The array C is going to store last
		// row of pascal triangle at the end.
		// And last entry of last row is nCr
		int C[] = new int[r + 1];

		C[0] = 1; // Top row of Pascal Triangle

		// One by constructs remaining rows of Pascal
		// Triangle from top to bottom
		for (int i = 1; i <= n; i++) {

			// Fill entries of current row using previous
			// row values
			for (int j = Math.min(i, r); j > 0; j--)

				// nCj = (n-1)Cj + (n-1)C(j-1);
				C[j] = (C[j] + C[j - 1]) % p;
		}
		return C[r];
	}

	static class Scanner {
		StringTokenizer st;
		BufferedReader br;

		public Scanner(InputStream s) {
			br = new BufferedReader(new InputStreamReader(s));
		}

		public Scanner(String file) throws Exception {
			br = new BufferedReader(new FileReader(file));
		}

		public int[] intArr(int n) throws IOException {
			int a[] = new int[n];
			for (int i = 0; i < a.length; i++) {
				a[i] = nextInt();
			}
			return a;
		}

		public long[] longArr(int n) throws IOException {
			long a[] = new long[n];
			for (int i = 0; i < a.length; i++) {
				a[i] = nextLong();
			}
			return a;
		}

		public String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}

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

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

		public String nextLine() throws IOException {
			return br.readLine();
		}

		public double nextDouble() throws IOException {
			String x = next();
			StringBuilder sb = new StringBuilder("0");
			double res = 0, f = 1;
			boolean dec = false, neg = false;
			int start = 0;
			if (x.charAt(0) == '-') {
				neg = true;
				start++;
			}
			for (int i = start; i < x.length(); i++)
				if (x.charAt(i) == '.') {
					res = Long.parseLong(sb.toString());
					sb = new StringBuilder("0");
					dec = true;
				} else {
					sb.append(x.charAt(i));
					if (dec)
						f *= 10;
				}
			res += Long.parseLong(sb.toString()) / f;
			return res * (neg ? -1 : 1);
		}

		public boolean ready() throws IOException {
			return br.ready();
		}

	}

	public static void shuffle(int[] a) {
		int n = a.length;
		for (int i = 0; i < n; i++) {
			int r = i + (int) (Math.random() * (n - i));
			int tmp = a[i];
			a[i] = a[r];
			a[r] = tmp;
		}
	}

}