A. Inna and Pink Pony----暴力

A. Inna and Pink Pony

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Dima and Inna are doing so great! At the moment, Inna is sitting on the magic lawn playing with a pink pony. Dima wanted to play too. He brought an n × m chessboard, a very tasty candy and two numbers a and b.

Dima put the chessboard in front of Inna and placed the candy in position (i, j) on the board. The boy said he would give the candy if it reaches one of the corner cells of the board. He's got one more condition. There can only be actions of the following types:

• move the candy from position (x, y) on the board to position (x - a, y - b);
• move the candy from position (x, y) on the board to position (x + a, y - b);
• move the candy from position (x, y) on the board to position (x - a, y + b);
• move the candy from position (x, y) on the board to position (x + a, y + b).

Naturally, Dima doesn't allow to move the candy beyond the chessboard borders.

Inna and the pony started shifting the candy around the board. They wonder what is the minimum number of allowed actions that they need to perform to move the candy from the initial position (i, j) to one of the chessboard corners. Help them cope with the task!

Input

The first line of the input contains six integers n, m, i, j, a, b (1 ≤ n, m ≤ 106; 1 ≤ i ≤ n; 1 ≤ j ≤ m; 1 ≤ a, b ≤ 106).

You can assume that the chessboard rows are numbered from 1 to n from top to bottom and the columns are numbered from 1 to m from left to right. Position (i, j) in the statement is a chessboard cell on the intersection of the i-th row and the j-th column. You can consider that the corners are: (1, m), (n, 1), (n, m), (1, 1).

Output

In a single line print a single integer — the minimum number of moves needed to get the candy.

If Inna and the pony cannot get the candy playing by Dima's rules, print on a single line "Poor Inna and pony!" without the quotes.

Examples

input

5 7 1 3 2 2

output

2

input

5 5 2 3 1 1

output

Poor Inna and pony!

Note

Note to sample 1:

Inna and the pony can move the candy to position (1 + 2, 3 + 2) = (3, 5), from there they can move it to positions(3 - 2, 5 + 2) = (1, 7) and (3 + 2, 5 + 2) = (5, 7). These positions correspond to the corner squares of the chess board. Thus, the answer to the test sample equals two.

題目鏈接：http://codeforces.com/contest/374/problem/A

這個題的題意真是絕了。

題意（轉）：有一個n*m的方格，位於(i,j)位置上，每次能同時向上下方向移動a並且向左右方向移動b（必須同時向兩個方向移動），求到達任意一個角上所需要的最小移動次數（（1，1），（1，m），（n，1），（n，m）中的任意一個），當然，任何時候不可以移動到方格外面

直觀的想法是求起始點到四個點的座標偏差，並計算x/y方向的運動次數，倘若兩者運動次數一樣，則可作爲一個解，

若運動次數奇偶性不同，則無法修正。遍歷四個點，取最小即可。

代碼：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;
int main(){
    int n,m,i,j,a,b,ans=1000000,x,y,ta,tb,sx[2],sy[2];
	scanf("%d%d%d%d%d%d",&n,&m,&i,&j,&a,&b);
	sx[0]=1,sx[1]=n,sy[0]=1,sy[1]=m;
	if((i==1||i==n)&&(j==1||j==m))
		ans=0;
    for(int p=0;p<2;p++){
      for(int q=0;q<2;q++){
		  x=abs(i-sx[p]);
		  y=abs(j-sy[q]);
		  if(x%a==0&&y%b==0){
             ta=x/a;
			 tb=y/b;
			 if(ta%2==tb%2){
			   if(ta==0&&(abs(i-1)<a)&&(abs(i-n)<a))
				   continue;
			   if(tb==0&&(abs(j-1)<b)&&(abs(j-m)<b))
				   continue;
			   ans=min(ans,max(ta,tb));
			 }
		  }
	  }
	}
	if(ans!=1000000)
    	printf("%d\n",ans);
	else
		printf("Poor Inna and pony!\n");
	return 0;
}