Alice has a cute cat. To keep her cat fit, Alice wants to design an exercising walk for her cat!

Initially, Alice’s cat is located in a cell (x,y) of an infinite grid. According to Alice’s theory, cat needs to move:

exactly a steps left: from (u,v) to (u?1,v);
exactly b steps right: from (u,v) to (u+1,v);
exactly c steps down: from (u,v) to (u,v?1);
exactly d steps up: from (u,v) to (u,v+1).
Note that the moves can be performed in an arbitrary order. For example, if the cat has to move 1 step left, 3 steps right and 2 steps down, then the walk right, down, left, right, right, down is valid.

Alice, however, is worrying that her cat might get lost if it moves far away from her. So she hopes that her cat is always in the area [x1,x2]×[y1,y2], i.e. for every cat’s position (u,v) of a walk x1≤u≤x2 and y1≤v≤y2 holds.

Also, note that the cat can visit the same cell multiple times.

Can you help Alice find out if there exists a walk satisfying her wishes?

Formally, the walk should contain exactly a+b+c+d unit moves (a to the left, b to the right, c to the down, d to the up). Alice can do the moves in any order. Her current position (u,v) should always satisfy the constraints: x1≤u≤x2, y1≤v≤y2. The staring point is (x,y).

You are required to answer t test cases independently.

Input
The first line contains a single integer t (1≤t≤103) — the number of testcases.

The first line of each test case contains four integers a, b, c, d (0≤a,b,c,d≤108, a+b+c+d≥1).

The second line of the test case contains six integers x, y, x1, y1, x2, y2 (?108≤x1≤x≤x2≤108, ?108≤y1≤y≤y2≤108).

Output
For each test case, output “YES” in a separate line, if there exists a walk satisfying her wishes. Otherwise, output “NO” in a separate line.

You can print each letter in any case (upper or lower).

Example
Input
6
3 2 2 2
0 0 -2 -2 2 2
3 1 4 1
0 0 -1 -1 1 1
1 1 1 1
1 1 1 1 1 1
0 0 0 1
0 0 0 0 0 1
5 1 1 1
0 0 -100 -100 0 100
1 1 5 1
0 0 -100 -100 100 0
Output
Yes
No
No
Yes
Yes
Yes
Note
In the first test case, one valid exercising walk is
(0,0)→(?1,0)→(?2,0)→(?2,1)→(?2,2)→(?1,2)→(0,2)→(0,1)→(0,0)→(?1,0)
思路：雖然是個(gè)A題，但是感覺還是挺有意義的。
不要考慮的太麻煩，只要從左右和上下兩方面來考慮就行。最貪心的方法就是左一下，右一下，直到某個(gè)動(dòng)作不能繼續(xù)，然后再執(zhí)行剩下的那個(gè)動(dòng)作。這樣我們就可以知道能不能實(shí)現(xiàn)了（比較距離）。上下也是一樣的道理。注意活動(dòng)范圍是一個(gè)點(diǎn)的情況。
代碼如下：

#include<bits/stdc++.h> #define ll long long using namespace std;int a,b,c,d; ll x,y,x1,y11,x2,y2;int main() {int t;scanf("%d",&t);while(t--){cin>>a>>b>>c>>d;cin>>x>>y>>x1>>y11>>x2>>y2;ll _min1=abs(b-a);ll _min2=abs(d-c);int flag=1;if(x2-x==0&&x1-x==0&&(a||b)) flag=0;if(y11-y==0&&y2-y==0&&(c||d)) flag=0;if(b>a&&x2-x<_min1) flag=0;if(b<a&&x-x1<_min1) flag=0;if(d<c&&y-y11<_min2) flag=0;if(d>c&&y2-y<_min2) flag=0;if(flag) cout<<"YES"<<endl;else cout<<"NO"<<endl;}return 0; }

努力加油a啊，(^o)/~

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