A. Inna and Pink Pony

time limit per test1 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.

 #include <iostream>
#include <cstdio>
#include <cmath>
using namespace std; int n, m, sx, sy, a, b; int main()
{
while(scanf("%d%d%d%d%d%d", &n, &m, &sx, &sy, &a, &b) != EOF)
{
int ans;
if((sx==&&sy==)||(sx==&&sy==m)||(sx==n&&sy==)||(sx==n&&sy==m))
{
printf("0\n");
}else if((n-)<a || (m-)<b)
printf("Poor Inna and pony!\n");
else
{
int a1 = (sx-)/a;
int a2 = (sy-)/b;
int a3 = (m-sy)/b;
int a4 = (n-sx)/a;
bool fg = false;
if(a1*a==(sx-) && a2*b==(sy-))
{
int tmp = abs(a1-a2);
if(tmp%==)
{
ans = max(a1, a2);
fg = true;
}
}
if(a1*a==(sx-) && a3*b==(m-sy))
{
int tmp = abs(a1-a3);
if(tmp%== && max(a1, a3) < ans)
{
ans = max(a1, a3);
fg = true;
}
}
if(a2*b==(sy-) && a4*a==(n-sx))
{
int tmp = abs(a4-a2);
if(tmp%== && max(a4, a2) < ans)
{
ans = max(a4, a2);
fg = true;
}
}
if(a3*b==(m-sy) && a4*a==(n-sx))
{
int tmp = abs(a3-a4);
if(tmp%== && max(a3, a4) < ans)
{
ans = max(a3, a4);
fg = true;
}
}
if(fg)printf("%d\n", ans);
else printf("Poor Inna and pony!\n");
}
} return ;
}

Codeforces374A的更多相关文章

随机推荐

  1. 关于textarea的应用--onchage,onpropertychange,oninput

    oninput,onpropertychange,onchange的用法 1.onchange触发事件必须满足两个条件: a)当前对象属性改变,并且是由键盘或鼠标事件激发的(脚本触发无效) b)当前对 ...

  2. 用Quick Cocos2dx做一个连连看(三)

    做个日记吧. 最近比较忙,斗志也不高,昨天有点时间了,开始做了一下连接方法,一开始用的搜索算法,但是bug比较多,究其原因是对语法和算法都不是很熟悉. 然后昨天下午利用点时间用稍微通俗一点的连接算法, ...

  3. 无法访问 ASP 兼容性模式

    <%@ Page Title="" Language="C#" MasterPageFile="../theme/classic/content ...

  4. 苹果App Store开发者帐户从申请,验证,到发布应用(1)

    app store为开发者提供四种类型的申请: 个人ios开发者计划$99/年 公司ios开发者计划$99/年 企业ios开发者计划$299/年 高校ios开发者计划免费 在这里主要介绍一下公司ios ...

  5. Maven的安装环境配置

    一.Maven的安装 二.Maven的配置 Settings.xml可以用来定义本地仓库.远程仓库.联网代理 Settings.xml文件可以存在两个地方: 1.多用户情况 conf目录下 2.单用户 ...

  6. Memcached源码分析之内存管理

    先再说明一下,我本次分析的memcached版本是1.4.20,有些旧的版本关于内存管理的机制和数据结构与1.4.20有一定的差异(本文中会提到). 一)模型分析在开始解剖memcached关于内存管 ...

  7. UVa 374 - Big Mod

    题目大意:计算R = BP mod M,根据模运算的性质计算. 正常计算会超时,可以用分治的思想降低时间复杂度.不过如果遇到00,结果...话说00的结果是1吗?忘了都... #include < ...

  8. 4)Javascript设计模式:Decorator模式

    function MacBook() { this.cost = function() { return 997; } } var macbook = new MacBook(); function ...

  9. Django 自定义模版标签和过滤器

    实现自定义过滤器 1. 创建register变量 在你的模块文件中,你必须首先创建一个全局register变量,它是用来注册你自定义标签和过滤器的, 你需要在你的python文件的开始处,插入几下代码 ...

  10. 关键词匹配(Ac自动机模板题)

    2772: 关键词匹配 Time Limit: 1 Sec  Memory Limit: 128 MBSubmit: 10  Solved: 4[Submit][Status][Web Board] ...