cf(旋转矩阵)

Description

Inna and Dima decided to surprise Sereja. They brought a really huge candy matrix, it's big even for Sereja! Let's number the rows of the giant matrix from 1 to n from
top to bottom and the columns — from 1 to m, from left to right. We'll represent the cell on the intersection of the i-th
row and j-th column as (i, j). Just as is expected, some cells of the giant candy matrix contain candies. Overall the
matrix has p candies: the k-th candy is at cell (xk, yk).

The time moved closer to dinner and Inna was already going to eat p of her favourite sweets from the matrix, when suddenly Sereja (for the reason he didn't share with anyone) rotated the matrix x times
clockwise by 90 degrees. Then he performed the horizontal rotate of the matrix y times. And then he rotated the matrix z times
counterclockwise by 90 degrees. The figure below shows how the rotates of the matrix looks like.



Inna got really upset, but Duma suddenly understood two things: the candies didn't get damaged and he remembered which cells contained Inna's favourite sweets before Sereja's strange actions. Help guys to find the new coordinates in the candy matrix after
the transformation Sereja made!

Input

The first line of the input contains fix integers n, m, x, y, z, p(1 ≤ n, m ≤ 109; 0 ≤ x, y, z ≤ 109; 1 ≤ p ≤ 105).

Each of the following p lines contains two integers xk, yk(1 ≤ xk ≤ n; 1 ≤ yk ≤ m) —
the initial coordinates of the k-th candy. Two candies can lie on the same cell.

Output

For each of the p candies, print on a single line its space-separated new coordinates.

Sample Input

Input

3 3 3 1 1 9
1 1
1 2
1 3
2 1
2 2
2 3
3 1
3 2
3 3

Output

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

Hint

Just for clarity. Horizontal rotating is like a mirroring of the matrix. For matrix:

QWER      REWQ
ASDF  ->  FDSA
ZXCV      VCXZ

题意是顺时针旋转矩阵x次，然后将矩阵做镜子一样的变换y次然后逆时针旋转z次之后求点对应的坐标。

需要我们找到旋转和翻转之后的坐标关系。

还有要注意的是旋转90度之后的m和n会对应的变换！而且变换了之后还不可以影响下一组数据，卡这两点真实卡死我了。。。。

#include <iostream>
#include <stdio.h>
#include <math.h>
#include<algorithm>
using namespace std;
int n, m, x, y, z, p;
int hx,hy;
void hh()
{
int a1=m,a2=n;
int dx=hx-1,dy=hy-1;
int xx=x%4,yy=y%2,zz=z%4;
if(xx==1)
hx=1+dy,hy=n-dx,swap(m,n);
else if(xx==2)
hx=n-dx,hy=m-dy;
else if(xx==3)
hx=m-dy,hy=1+dx,swap(m,n);
dx=hx-1,dy=hy-1;
if(yy==1)
hx=1+dx,hy=m-dy;
dx=hx-1,dy=hy-1;
if(zz==1)
hx=m-dy,hy=1+dx,swap(m,n);
else if(zz==2)
hx=n-dx,hy=m-dy;
else if(zz==3)
hx=1+dy,hy=n-dx,swap(m,n);
m=a1,n=a2;
cout<<hx<<' '<<hy<<endl;
}
int main()
{
cin>>n>>m>>x>>y>>z>>p;
for(int i=0;i<p;i++)
{
scanf("%d%d",&hx,&hy);
hh();
}
return 0;
}