Codeforces 1315A Dead Pixel （水题）

Screen resolution of Polycarp's monitor is a×ba×b pixels. Unfortunately, there is one dead pixel at his screen. It has coordinates (x,y)(x,y) (0≤x<a,0≤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≤t≤1041≤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≤a,b≤1041≤a,b≤104 ; 0≤x<a0≤x<a ; 0≤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.

Example

Input

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

Output

56
6
442
1
45
80
找出这个点左右上下四部分最大的一部分即可。

#include <bits/stdc++.h>
using namespace std;
int main()
{
    int t;
    cin>>t;
    int a,b,x,y;
    while(t--)
    {
        cin>>a>>b>>x>>y;
        int s1=a*y;
        int s2=b*x;
        int s3=a*(b-y-);
        int s4=b*(a-x-);
        s2=max(s1,s2);
        s4=max(s3,s4);
        cout<<max(s2,s4)<<endl;
    }
    return ;
}