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:

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Solution (in CPP):

Solution:
#include <bits/stdc++.h>
const int MAX = 100001;
const int MOD = 1e9 + 7;
int ara[MAX];
using namespace std;
void fast() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
}
long long gcd(long long a, long long b) {
  if (b == 0)
    return a;
  else
    gcd(b, a % b);
}
long long lcd(long long a, long long b) { return (a * b) / gcd(a, b); }
bool f(long long a, long long b) { return a > b; }
long long max_sum_subarrays(long long ara[], long long n) {
  long long ans = 0, sum = 0;
  for (long long k = 0; k < n; k++) {
    sum = max(ara[k], sum + ara[k]);
    ans = max(ans, sum);
  }
  return ans;
}
long long num_divisors(long long n) {
  long long cnt = 0;
  for (int i = 1; i * i <= n; i++) {
    if (n % i == 0 && i * i == n) cnt += 1;
    if (n % i == 0 && i * i != n) cnt += 2;
  }
  return cnt;
}
long long comb(long long m, long long n) {
  long long x = max(m - n, n), ans = 1, i = 1;
  for (long long k = m; k >= x + 1; k--) {
    ans *= k;
    ans /= i;
    i++;
  }
  return ans;
}
long long binsearch(long long low, long long high, long long target) {
  while (low <= high) {
    long long mid = (low + high) / 2;
    if (ara[mid] > target)
      high = mid - 1;
    else if (ara[mid] < target)
      low = mid + 1;
    else
      return mid;
  }
  return -1;
}
void solve() {
  long long n, m, x, y;
  cin >> n >> m >> x >> y;
  x -= 1;
  y -= 1;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      cout << x + 1 << " " << y + 1 << "\n";
      if (j != m - 1) y = (y + 1) % m;
    }
    x = (x + 1) % n;
  }
}
int main() {
  fast();
  solve();
  return 0;
}