You have a board represented as a grid with 2 × 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 × 1 tiles, both cells are colored in white) and b black dominoes (2 × 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

The first line contains a single integer t (1 ≤ t ≤ 3000) — the number of test cases.

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

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

Output

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).

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


Output


NO
YES
NO
YES
YES

Note

In the first test case, n = 1, k_1 = 0 and k_2 = 1. It means that 2 × 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 × 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 × 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)).
from collections import defaultdict,OrderedDict,Counter
from sys import stdin,stdout
from bisect import bisect_left,bisect_right
# import numpy as np
from queue import Queue,PriorityQueue
from heapq import *
from statistics import *
from math import *
import fractions
import copy
from copy import deepcopy
import sys
import io
sys.setrecursionlimit(10000)
import math
import os
import bisect
import collections
mod=pow(10,9)+7
import random
from random import *
from time import  time;
def ncr(n, r, p=mod):
    num = den = 1
    for i in range(r):
        num = (num * (n - i)) % p
        den = (den * (i + 1)) % p
    return (num * pow(den,
            p - 2, p)) % p
def normalncr(n,r):
    r=min(r,n-r)
    count=1;
    for i in range(n-r,n+1):
        count*=i;

    for i in range(1,r+1):
        count//=i;
    return count

inf=float("inf")
adj=defaultdict(set)
visited=defaultdict(int)
def addedge(a,b):
    adj[a].add(b)
    adj[b].add(a)
def bfs(v):
    q=Queue()
    q.put(v)
    visited[v]=1
    while q.qsize()>0:
        s=q.get_nowait()
        print(s)
        for i in adj[s]:
            if visited[i]==0:
                q.put(i)
                visited[i]=1
def dfs(v,visited):
    if visited[v]==1:
        return;
    visited[v]=1

    print(v)
    for i in adj[v]:
        dfs(i,visited)
# a9=pow(10,6)+10
# prime = [True for i in range(a9 + 1)]
# def SieveOfEratosthenes(n):
#     p = 2
#     while (p * p <= n):
#         if (prime[p] == True):
#             for i in range(p * p, n + 1, p):
#                 prime[i] = False
#         p += 1
# SieveOfEratosthenes(a9)
# prime_number=[]
# for i in range(2,a9):
#     if prime[i]:
#         prime_number.append(i)
def reverse_bisect_right(a, x, lo=0, hi=None):
    if lo < 0:
        raise ValueError('lo must be non-negative')
    if hi is None:
        hi = len(a)
    while lo < hi:
        mid = (lo+hi)//2
        if x > a[mid]:
            hi = mid
        else:
            lo = mid+1
    return lo
def reverse_bisect_left(a, x, lo=0, hi=None):
    if lo < 0:
        raise ValueError('lo must be non-negative')
    if hi is None:
        hi = len(a)
    while lo < hi:
        mid = (lo+hi)//2
        if x >= a[mid]:
            hi = mid
        else:
            lo = mid+1
    return lo

def get_list():
    return list(map(int,input().split()))
def get_str_list_in_int():
    return [int(i) for i in list(input())]
def get_str_list():
    return list(input())
def get_map():
    return map(int,input().split())
def input_int():
    return int(input())
def matrix(a,b):
    return [[0 for i in range(b)] for j in range(a)]
def swap(a,b):
    return b,a
def find_gcd(l):
    a=l[0]
    for i in range(len(l)):
        a=gcd(a,l[i])
    return a;
def is_prime(n):
    sqrta=int(sqrt(n))
    for i in range(2,sqrta+1):
        if n%i==0:
            return 0;
    return 1;
def prime_factors(n):
    while n % 2 == 0:
        return [2]+prime_factors(n//2)
    sqrta = int(sqrt(n))


    for i in range(3,sqrta+1,2):
        if n%i==0:
            return [i]+prime_factors(n//i)

    return [n]
def p(a):
    if type(a)==str:
        print(a+"\n")
    else:
        print(str(a)+"\n")
def ps(a):
    if type(a)==str:
        print(a)
    else:
        print(str(a))
def kth_no_not_div_by_n(n,k):
    return k+(k-1)//(n-1)
def forward_array(l):
    n=len(l)
    stack = []
    forward=[0]*n
    for i in range(len(l) - 1, -1, -1):
        while len(stack) and l[stack[-1]] < l[i]:
            stack.pop()
        if len(stack) == 0:
            forward[i] = len(l);
        else:
            forward[i] = stack[-1]
        stack.append(i)
    return forward;
def backward_array(l):
    n=len(l)
    stack = []
    backward=[0]*n
    for i in range(len(l)):
        while len(stack) and l[stack[-1]] < l[i]:
            stack.pop()
        if len(stack) == 0:
            backward[i] = -1;
        else:
            backward[i] = stack[-1]
        stack.append(i)
    return backward
nc="NO"
yc="YES"
ns="No"
ys="Yes"

# import math as mt
# MAXN=10**7
# spf = [0 for i in range(MAXN)]
# def sieve():
#     spf[1] = 1
#     for i in range(2, MAXN):
#         # marking smallest prime factor
#         # for every number to be itself.
#         spf[i] = i
#
#         # separately marking spf for
#     # every even number as 2
#     for i in range(4, MAXN, 2):
#         spf[i] = 2
#
#     for i in range(3, mt.ceil(mt.sqrt(MAXN))):
#
#         # checking if i is prime
#         if (spf[i] == i):
#
#             # marking SPF for all numbers
#             # divisible by i
#             for j in range(i * i, MAXN, i):
#
#                 # marking spf[j] if it is
#                 # not previously marked
#                 if (spf[j] == j):
#                     spf[j] = i
# def getFactorization(x):
#     ret = list()
#     while (x != 1):
#         ret.append(spf[x])
#         x = x // spf[x]
#
#     return ret
# sieve()






# input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
# input=stdin.readline
# print=stdout.write
t=int(input())
for i in range(t):
    n,k1,k2=get_map()
    w,b=get_map();
    white=min(k1,k2)
    black=n-max(k1,k2)

    white+=(abs(k1-k2))//2
    black+=abs(k1-k2)//2
    if w<=white and b<=black:
        print(yc)
    else:
        print(nc)