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#include <bits/stdc++.h>
using namespace std;
int n, m; //n*m的棋盘
int x, y; //马的位置
const int N = 410;
int ans[N][N]; //到任意点需要最少走几步
//坐标
struct coord {
int x, y;
};
queue<coord> q;
//马的八大方向坐标位移
int walk[8][2] = {{2, 1},
{1, 2},
{-1, 2},
{-2, 1},
{-2, -1},
{-1, -2},
{1, -2},
{2, -1}};
int main() {
//输入棋盘大小和马的位置
cin >> n >> m >> x >> y;
//标识为-1表示没有探索过
memset(ans, -1, sizeof ans);
//马入队列
q.push({x, y});
ans[x][y] = 0;//出发点标识为最少走0步
while (!q.empty()) {
coord u = q.front();
q.pop();
for (int k = 0; k < 8; k++) {//8个方向
int x = u.x + walk[k][0], y = u.y + walk[k][1];//可能达到的下一步坐标
if (x < 1 || x > n || y < 1 || y > m || ans[x][y] != -1)continue;
//步数等于上一级来的位置+1
ans[x][y] = ans[u.x][u.y] + 1;
//入队列
q.push({x, y});
}
}
//输出结果
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= m; j++)
printf("%-5d", ans[i][j]); //左对齐
puts("");
}
return 0;
}