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 CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 5;
int dir[4][2] = {{0, 1}, {0, -1}, {-1, 0}, {1, 0}};
long long gcd(long long n, long long m) { return m == 0 ? n : gcd(m, n % m); }
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
int day[13] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
int run(int y) {
  if (y % 400 == 0) {
    return 1;
  }
  if (y % 100 != 0 && y % 4 == 0) {
    return 1;
  }
  return 0;
}
long long q_p(long long x, long long n, long long mod) {
  long long ans = 1;
  while (n) {
    if (n & 1) {
      ans = x * ans % mod;
    }
    n = n >> 1;
    x = x * x % mod;
  }
  return ans;
}
long long qpow(long long x, long long n) {
  long long ans = 1;
  while (n) {
    if (n & 1) {
      ans = x * ans;
    }
    n = n >> 1;
    x = x * x;
  }
  return ans;
}
long long jc(long long n, long long cnt) {
  long long sum = 1;
  while (cnt != 0) {
    cnt--;
    sum *= n;
    n -= 1;
  }
  return sum;
}
int dp1() {
  int n, k;
  int d[100][100];
  cin >> n >> k;
  int v[n + 10], w[n + 10];
  for (int i = 1; i <= n; i++) {
    cin >> w[i];
    cin >> v[i];
  }
  for (int i = 0; i <= n; i++) {
    for (int j = 0; j < n; j++) {
      d[i][j] = 0;
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= k; j++) {
      if (j < w[i]) {
        d[i][j] = d[i - 1][j];
      } else {
        if (d[i - 1][j] > d[i - 1][j - w[i]] + v[i]) {
          d[i][j] = d[i - 1][j];
        } else {
          d[i][j] = d[i - 1][j - w[i]] + v[i];
        }
      }
    }
  }
  return d[n][k];
}
int prim[168] = {
    2,   3,   5,   7,   11,  13,  17,  19,  23,  29,  31,  37,  41,  43,
    47,  53,  59,  61,  67,  71,  73,  79,  83,  89,  97,  101, 103, 107,
    109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181,
    191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263,
    269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 349,
    353, 359, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421, 431, 433,
    439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499, 503, 509, 521,
    523, 541, 547, 557, 563, 569, 571, 577, 587, 593, 599, 601, 607, 613,
    617, 619, 631, 641, 643, 647, 653, 659, 661, 673, 677, 683, 691, 701,
    709, 719, 727, 733, 739, 743, 751, 757, 761, 769, 773, 787, 797, 809,
    811, 821, 823, 827, 829, 839, 853, 857, 859, 863, 877, 881, 883, 887,
    907, 911, 919, 929, 937, 941, 947, 953, 967, 971, 977, 983, 991, 997};
struct M {
  long long data[2][2];
  M() { memset(data, 0, sizeof(data)); }
};
M mul(M a, M b) {
  M ans;
  ans.data[0][0] = (a.data[0][0] * b.data[0][0] + a.data[0][1] * b.data[1][0]);
  ans.data[0][1] = (a.data[0][0] * b.data[0][1] + a.data[0][1] * b.data[1][1]);
  ans.data[1][0] = (a.data[1][0] * b.data[0][0] + a.data[1][1] * b.data[1][0]);
  ans.data[1][1] = (a.data[1][0] * b.data[0][1] + a.data[1][1] * b.data[1][1]);
  return ans;
}
M qqpow(M m, long long n) {
  M E;
  E.data[0][0] = 1;
  E.data[1][1] = 1;
  while (n != 0) {
    if (n & 1) {
      E = mul(E, m);
    }
    m = mul(m, m);
    n = n >> 1;
  }
  return E;
}
const int mm = 1e5 + 10;
vector<long long> p[mm];
int vis[mm];
long long w[mm], g[mm];
long long ww[mm];
long long ans = 0;
int n, k;
void dfs_s(int root, int cnt, int f) {
  if (vis[root]) {
    cnt++;
  }
  if (cnt > 2) {
    return;
  }
  if (cnt <= 2 && p[root].size() == 1) {
    ans++;
    return;
  }
  for (int i = 0; i < p[root].size(); i++) {
    int son = p[root][i];
    if (son != f) {
      dfs_s(son, cnt, root);
    }
  }
}
int h[100000], W[100000];
int book[510][510];
int H;
int hh;
inline int read() {
  int f = 1, num = 0;
  char ch = getchar();
  while (0 == isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (0 != isdigit(ch))
    num = (num << 1) + (num << 3) + ch - '0', ch = getchar();
  return num * f;
}
void ksort(int a[], int l, int r) {
  int i = l, j = r;
  int temp = a[i];
  while (i < j) {
    while (i < j && a[j] >= temp) {
      j--;
    }
    if (i < j) {
      a[i] = a[j];
      i++;
    }
    while (i < j && a[i] < temp) {
      i++;
    }
    if (i < j) {
      a[j] = a[i];
      j--;
    }
  }
  a[i] = temp;
  if (i > l) ksort(a, l, i - 1);
  if (i < r) ksort(a, i + 1, r);
}
int m;
int dx, dy;
int cx, cy;
char mp[1010][1010];
struct node {
  int jc[N], v1, v2;
  long long mod;
  void init() {
    jc[0] = 1;
    for (int i = 1; i < N; i++) jc[i] = 1ll * jc[i - 1] * i % mod;
  }
  void add1(int x) { v1 = 1ll * v1 * jc[x] % mod; }
  void add2(int x) { v2 = 1ll * v2 * jc[x] % mod; }
} rd, nd, cd, c[4];
queue<struct node> qq;
int bfs() {}
int main(int argc, char** argv) {
  int n, m, sx, sy;
  cin >> n >> m >> sx >> sy;
  cout << sx << " " << sy << endl;
  cout << 1 << " " << sy << endl;
  for (int i = 1; i <= n; i++) {
    if (i % 2 == 0) {
      for (int j = m; j >= 1; j--) {
        if (i == 1 && j == sy) {
          continue;
        }
        if (i == sx && j == sy) {
          continue;
        }
        cout << i << " " << j << endl;
      }
    } else {
      for (int j = 1; j <= m; j++) {
        if (i == 1 && j == sy) {
          continue;
        }
        if (i == sx && j == sy) {
          continue;
        }
        cout << i << " " << j << endl;
      }
    }
  }
  return 0;
}