Dead Pixel

Screen resolution of Polycarp's monitor is a \times ba×b pixels. Unfortunately, there is one dead pixel at his screen. It has coordinates (x, y)(x,y) (0 \le x < a, 0 \le y < b0≤x<a,0≤y<b). You can consider columns of pixels to be numbered from 00 to a-1a−1, and rows — from 00 to b-1b−1.

Polycarp wants to open a rectangular window of maximal size, which doesn't contain the dead pixel. The boundaries of the window should be parallel to the sides of the screen.

Print the maximal area (in pixels) of a window that doesn't contain the dead pixel inside itself.

Input

In the first line you are given an integer tt (1 \le t \le 10^41≤t≤104) — the number of test cases in the test. In the next lines you are given descriptions of tt test cases.

Each test case contains a single line which consists of 44 integers a, b, xa,b,x and yy (1 \le a, b \le 10^41≤a,b≤104; 0 \le x < a0≤x<a; 0 \le y < b0≤y<b) — the resolution of the screen and the coordinates of a dead pixel. It is guaranteed that a+b>2a+b>2 (e.g. a=b=1a=b=1 is impossible).

Output

Print tt integers — the answers for each test case. Each answer should contain an integer equal to the maximal possible area (in pixels) of a rectangular window, that doesn't contain the dead pixel.

Sample 1

Inputcopy	Outputcopy
6 8 8 0 0 1 10 0 3 17 31 10 4 2 1 0 0 5 10 3 9 10 10 4 8	56 6 442 1 45 80

Note

In the first test case, the screen resolution is 8 \times 88×8, and the upper left pixel is a dead pixel. Here you can see one of two possible layouts of the maximal window.

题意：

在一个axb的屏幕上有一个坏点（x,y），问不包含坏点的最大矩形面积。

思路：

比较可能四个矩形的大小。

代码：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<string>
#include<cstring>
#include<cmath>
#include<map>
#include<vector>
using namespace std;
typedef long long ll;
void slove(){int t;cin>>t;while(t--){int a,b,x,y;cin>>a>>b>>x>>y;int ans=a*y;ans=max(ans,b*x);ans=max(ans,a*(b-y-1));ans=max(ans,b*(a-x-1));cout<<ans<<endl;}
}
int main(){slove();
}

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