Domino on Windowsill
Contest: Educational Codeforces Round 106 (Rated for Div. 2)
Difficulty rating: 800
Tags: combinatorics, constructive algorithms, math

You have a board represented as a grid with $$$2 \times n$$$ cells.

The first $$$k_1$$$ cells on the first row and first $$$k_2$$$ cells on the second row are colored in white. All other cells are colored in black.

You have $$$w$$$ white dominoes ($$$2 \times 1$$$ tiles, both cells are colored in white) and $$$b$$$ black dominoes ($$$2 \times 1$$$ tiles, both cells are colored in black).

You can place a white domino on the board if both board's cells are white and not occupied by any other domino. In the same way, you can place a black domino if both cells are black and not occupied by any other domino.

Can you place all $$$w + b$$$ dominoes on the board if you can place dominoes both horizontally and vertically?

Input format:
The first line contains a single integer $$$t$$$ ($$$1 \le t \le 3000$$$) — the number of test cases.

The first line of each test case contains three integers $$$n$$$, $$$k_1$$$ and $$$k_2$$$ ($$$1 \le n \le 1000$$$; $$$0 \le k_1, k_2 \le n$$$).

The second line of each test case contains two integers $$$w$$$ and $$$b$$$ ($$$0 \le w, b \le n$$$).

Output format:
For each test case, print YES if it's possible to place all $$$w + b$$$ dominoes on the board and NO, otherwise.

You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES are all recognized as positive answer).

Constraints:
Time limit: 1.0 s
Memory limit: 256.0 MB

Note:
In the first test case, $$$n = 1$$$, $$$k_1 = 0$$$ and $$$k_2 = 1$$$. It means that $$$2 \times 1$$$ board has black cell $$$(1, 1)$$$ and white cell $$$(2, 1)$$$. So, you can't place any white domino, since there is only one white cell.

In the second test case, the board of the same size $$$2 \times 1$$$, but both cell are white. Since $$$w = 0$$$ and $$$b = 0$$$, so you can place $$$0 + 0 = 0$$$ dominoes on the board.

In the third test case, board $$$2 \times 3$$$, but fully colored in black (since $$$k_1 = k_2 = 0$$$), so you can't place any white domino.

In the fourth test case, cells $$$(1, 1)$$$, $$$(1, 2)$$$, $$$(1, 3)$$$, and $$$(2, 1)$$$ are white and other cells are black. You can place $$$2$$$ white dominoes at positions $$$((1, 1), (2, 1))$$$ and $$$((1, 2), (1, 3))$$$ and $$$2$$$ black dominoes at positions $$$((1, 4), (2, 4))$$$ $$$((2, 2), (2, 3))$$$.

Example input:
5
1 0 1
1 0
1 1 1
0 0
3 0 0
1 3
4 3 1
2 2
5 4 3
3 1

Example output:
NO
YES
NO
YES
YES

Editorial:
We can prove that if we have $$$k_1 + k_2$$$ white cells on the board then we can place any $$$w$$$ white dominoes as long as $$$2w \le k_1 + k_2$$$.
The proof is the following: if $$$k_1 \ge k_2$$$ let's place one domino at position $$$((1, k_1 - 1), (1, k_1))$$$, otherwise let's place domino at position $$$((2, k_2 - 1), (2, k_2))$$$. Then we can solve the placement of $$$w - 1$$$ dominoes in $$$k_1 - 2$$$ cells in the first row and $$$k_2$$$ cells of the second row recursively (or, analogically, $$$k_1$$$ and $$$k_2 - 2$$$).
At the end, either all dominoes are placed or $$$k_1 < 2$$$ and $$$k_2 < 2$$$. If $$$k_1 = 0$$$ or $$$k_2 = 0$$$ then, since $$$2w \le k_1 + k_2$$$, then $$$w = 0$$$ or we successfully placed all dominoes. If $$$k_1 = 1$$$ and $$$k_2 = 1$$$ then we, possibly, need to place one domino more — and we can place it vertically.
We can prove that we can place any $$$b$$$ dominoes as long as $$$2b \le (n - k_1) + (n - k_2)$$$ in the same manner.
As a result, all we need to check is that $$$2w \le k_1 + k_2$$$ and $$$2b \le (n - k_1) + (n - k_2)$$$.