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>
using namespace std;
long long expo(long long a, long long b, long long m) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a) % m;
    a = (a * a) % m;
    b = b >> 1;
  }
  return res;
}
long long mminvprime(long long a, long long b) { return expo(a, b - 2, b); }
long long mod_add(long long a, long long b, long long m) {
  a = a % m;
  b = b % m;
  return (((a + b) % m) + m) % m;
}
long long mod_mul(long long a, long long b, long long m) {
  a = a % m;
  b = b % m;
  return (((a * b) % m) + m) % m;
}
long long mod_sub(long long a, long long b, long long m) {
  a = a % m;
  b = b % m;
  return (((a - b) % m) + m) % m;
}
long long mod_div(long long a, long long b, long long m) {
  a = a % m;
  b = b % m;
  return (mod_mul(a, mminvprime(b, m), m) + m) % m;
}
vector<int> factors(int x) {
  vector<int> result;
  int i = 1;
  while (i * i <= x) {
    if (x % i == 0) {
      result.push_back(i);
      if (x / i != i) {
        result.push_back(x / i);
      }
    }
    i++;
  }
  return result;
}
bool compare(const pair<int, int>& i, const pair<int, int>& j) {
  return i.second < j.second;
}
long long pow(long long a, long long b, long long m) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = (ans * a) % m;
    b /= 2;
    a = (a * a) % m;
  }
  return ans;
}
vector<long long> pr;
void sieve(long long n) {
  bool prime[n + 1];
  memset(prime, true, sizeof(prime));
  for (long long p = 2; p * p <= n; p++) {
    if (prime[p] == true) {
      for (long long i = p * p; i <= n; i += p) prime[i] = false;
    }
  }
  for (long long p = 2; p <= n; p++)
    if (prime[p]) pr.push_back(p);
}
int gcd(int a, int b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
void leftRotate(int arr[], int d, int n) {
  d = d % n;
  int g_c_d = gcd(d, n);
  for (int i = 0; i < g_c_d; i++) {
    int temp = arr[i];
    int j = i;
    while (1) {
      int k = j + d;
      if (k >= n) k = k - n;
      if (k == i) break;
      arr[j] = arr[k];
      j = k;
    }
    arr[j] = temp;
  }
}
int main() {
  int t;
  t = 1;
  while (t--) {
    int n, m;
    cin >> n >> m;
    int x, y;
    cin >> x >> y;
    for (int i = 1; i <= n; i++) {
      if (i % 2 != 0) {
        for (int j = 1; j <= m; j++) {
          int tmp1 = (i + x - 2) % n + 1;
          int tmp2 = (j + y - 2) % m + 1;
          cout << tmp1 << " " << tmp2 << endl;
        }
      } else {
        for (int j = m; j >= 1; j--) {
          int tmp1 = (i + x - 2) % n + 1;
          int tmp2 = (j + y - 2) % m + 1;
          cout << tmp1 << " " << tmp2 << endl;
        }
      }
    }
  }
  return 0;
}