E. Working routine

time limit per test

2.5 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Vasiliy finally got to work, where there is a huge amount of tasks waiting for him. Vasiliy is given a matrix consisting of n rows and m

columns and q tasks. Each task is to swap two submatrices of the given matrix.

For each task Vasiliy knows six integers ai, bi, ci, di, hi, wi, where ai is the index of the row where the top-left corner of the first rectangle is located, bi is the index of its column, c

i is the index of the row of the top-left corner of the second rectangle, di is the index of its column, hi is the height of the rectangle and wi is its width.

It's guaranteed that two rectangles in one query do not overlap and do not touch, that is, no cell belongs to both rectangles, and no two cells belonging to different rectangles share a side. However, rectangles are allowed to share an angle.

Vasiliy wants to know how the matrix will look like after all tasks are performed.

Input

The first line of the input contains three integers n, m and q (2 ≤ n, m ≤ 1000, 1 ≤ q ≤ 10 000) — the number of rows and columns in matrix, and the number of tasks Vasiliy has to perform.

Then follow n lines containing m integers vi, j (1 ≤ vi, j ≤ 109) each — initial values of the cells of the matrix.

Each of the following q lines contains six integers ai, bi, ci, di, hi, wi (1 ≤ ai, ci, hi ≤ n, 1 ≤ bi, di, wi ≤ m).

Output

Print n lines containing m integers each — the resulting matrix.

Examples

input

Copy

4 4 2
1 1 2 2
1 1 2 2
3 3 4 4
3 3 4 4
1 1 3 3 2 2
3 1 1 3 2 2

output

Copy

4 4 3 3
4 4 3 3
2 2 1 1
2 2 1 1

input

Copy

4 2 1
1 1
1 1
2 2
2 2
1 1 4 1 1 2

output

Copy

2 2
1 1
2 2
1 1

題意：給一個矩陣，有q次操作，每次交換兩個不相交的子矩陣

 

可以發現，每次交換時子矩陣內部位置關係是不變的，所以只要交換兩個子矩陣的外部位置就行了，每個點開一個四個方向的連結串列，存每個方向的點，每次交換隻要交換連結串列即可，具體看程式碼

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
const int maxm = 1005;
struct node
{
	int x, y;
	node() {};
	node(int x, int y) :x(x), y(y) {};
}U[maxm][maxm], D[maxm][maxm], L[maxm][maxm], R[maxm][maxm];
int f[maxm][maxm];
int main()
{
	int n, i, j, k, sum, m, Q;
	int a, b, c, d, h, w;
	node u, v, p, q;
	scanf("%d%d%d", &n, &m, &Q);
	for (i = 1;i <= n;i++)
		for (j = 1;j <= m;j++)
			scanf("%d", &f[i][j]);
	for (i = 0;i <= n + 1;i++)
	{
		for (j = 0;j <= m + 1;j++)
		{
			U[i][j] = node(i - 1, j), D[i][j] = node(i + 1, j);
			L[i][j] = node(i, j - 1), R[i][j] = node(i, j + 1);
		}
	}
	while (Q--)
	{
		scanf("%d%d%d%d%d%d", &a, &b, &c, &d, &h, &w);
		u.x = a, u.y = 0, v.x = c, v.y = 0;
		for (i = 1;i <= b;i++) u = R[u.x][u.y];
		for (i = 1;i <= d;i++) v = R[v.x][v.y];
		for (i = 1;i <= w;i++)
		{
			p = U[u.x][u.y], q = U[v.x][v.y];
			U[u.x][u.y] = q, U[v.x][v.y] = p;
			D[p.x][p.y] = v, D[q.x][q.y] = u;
			if (i < w) u = R[u.x][u.y], v = R[v.x][v.y];
		}
		for (i = 1;i <= h;i++)
		{
			p = R[u.x][u.y], q = R[v.x][v.y];
			R[u.x][u.y] = q, R[v.x][v.y] = p;
			L[p.x][p.y] = v, L[q.x][q.y] = u;
			if (i < h) u = D[u.x][u.y], v = D[v.x][v.y];
		}
		for (i = 1;i <= w;i++)
		{
			p = D[u.x][u.y], q = D[v.x][v.y];
			D[u.x][u.y] = q, D[v.x][v.y] = p;
			U[p.x][p.y] = v, U[q.x][q.y] = u;
			if (i < w) u = L[u.x][u.y], v = L[v.x][v.y];
		}
		for (i = 1;i <= h;i++)
		{
			p = L[u.x][u.y], q = L[v.x][v.y];
			L[u.x][u.y] = q, L[v.x][v.y] = p;
			R[p.x][p.y] = v, R[q.x][q.y] = u;
			if (i < h) u = U[u.x][u.y], v = U[v.x][v.y];
		}
	}
	for (i = 1;i <= n;i++)
	{
		p.x = i, p.y = 0;
		for (j = 1;j <= m;j++)
		{
			p = R[p.x][p.y];
			printf("%d%c", f[p.x][p.y], j == m ? '\n' : ' ');
		}
	}
	return 0;
}