B. Boboniu Plays Chess

Problem - 1395B - Codeforces

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×mn×m grid (rows are numbered from 11 to nn, and columns are numbered from 11 to mm). You have a chess piece, and it stands at some cell (Sx,Sy)(Sx,Sy) which is not on the border (i.e. 2≤Sx≤n−12≤Sx≤n−1 and 2≤Sy≤m−12≤Sy≤m−1).

From the cell (x,y)(x,y), you can move your chess piece to (x,y′)(x,y′) (1≤y′≤m,y′≠y1≤y′≤m,y′≠y) or (x′,y)(x′,y) (1≤x′≤n,x′≠x1≤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 nn, mm, SxSx and SySy (3≤n,m≤1003≤n,m≤100, 2≤Sx≤n−12≤Sx≤n−1, 2≤Sy≤m−12≤Sy≤m−1) — the number of rows, the number of columns, and the initial position of your chess piece, respectively.

Output

You should print n⋅mn⋅m lines.

The ii-th line should contain two integers xixi and yiyi (1≤xi≤n1≤xi≤n, 1≤yi≤m1≤yi≤m), denoting the ii-th cell that you visited. You should print exactly nmnm pairs (xi,yi)(xi,yi), they should cover all possible pairs (xi,yi)(xi,yi), such that 1≤xi≤n1≤xi≤n, 1≤yi≤m1≤yi≤m.

We can show that under these constraints there always exists a solution. If there are multiple answers, print any.

Examples

input

Copy

3 3 2 2

output

Copy

2 2
1 2
1 3
2 3
3 3
3 2
3 1
2 1
1 1

input

Copy

3 4 2 2

output

Copy

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

题意：从sx，sy点出顺着(x′,y)或者(x,y′)遍历打印所有路径

Possible routes for two examples:

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
// ctrl+shift+C 注释
//ctrl+shift+x 取消
#define int long long
#define YES cout<<"YES"<
#define NO cout<<"NO"<
#define fast ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
typedef long long ll;
typedef pair<int,int> PII;
const int N=2e5+10;
const ll M=1e18+10;
const int mod=1e9+7;
int a[N],sum[N];
priority_queue<int,vector<int>,greater<int> >pq;
set<int>se;
map<int,int>mp;
queue<int>qu;
vector<int>v;
deque<int>de;
int n,m,sx,sy;
void solve()
{
    cin>>n>>m>>sx>>sy;
    int x=sx,y=sy,id=sy;
    for(int i=1;i<=n*m;i++)
    {
       while(x>=1)
       {
           for(int i=y;i<=m;i++)
           {
               cout<" "<
               id=i;
           }
           for(int i=y-1;i>=1;i--)
           {
               cout<" "<
               id=i;
           }
           y=id;
           x--;
       }
       sy=id;
      while(sx
       {
            sx++;
           for(int i=sy;i<=m;i++)
           {
               cout<" "<
               id=i;
           }
           for(int i=sy-1;i>=1;i--)
           {
               cout<" "<
               id=i;
           }
           sy=id;
       }
    }
}
signed main()
{
   int t=1;
   //cin>>t;
   while(t--)
   {
       solve();
   }
}