Problem: The progress is not standing still in Berland. Recently all garbage containers in Bertown, the capital of Berland, were replaced by differentiated recycling bins, each accepting some category of waste. While this will definitely improve the ecological situation, for some citizens it's difficult to get used to the habit of sorting waste.

Monocarp is one of those citizens who tries to get used to waste sorting. Today he has to take out the trash from his house. There are three containers near the Monocarp's house, the first one accepts paper waste, the second one accepts plastic waste, and the third one — all other types of waste. It is possible to fit c_1 items into the first container, c_2 items into the second container and c_3 items into the third container.

Monocarp has a lot of items to throw into containers. Some are made of paper, so Monocarp has to put them into the first container (he has a_1 such items), some are made of plastic, so he has to put them into the second container (he has a_2 such items), and some are neither paper nor plastic — so Monocarp has to put them into the third container (he has a_3 such items).

Unfortunately, there are also two categories of items that Monocarp is unsure of: he has a_4 items which are partially made of paper, so he will put each of these items either into the first container or into the third container. Similarly, he has a_5 items partially made of plastic, so he has to put each of them either into the second container or into the third container. Obviously, this choice is made separately for each item — for example, Monocarp can throw several partially-plastic items into the second container, and all other partially-plastic items — into the third one.

Now Monocarp wonders: is it possible to put each item into some container so that the first container will hold no more than c_1 items, the second one — no more than c_2 items, and the third one — no more than c_3 items?

Input

The first line contains one integer t (1 ≤ t ≤ 3 ⋅ 10^4) — the number of test cases. 

Each test case consists of two lines. The first line of each test case contains three integers c_1, c_2, c_3 (0 ≤ c_1, c_2, c_3 ≤ 10^8) — the capacities of the containers. 

The second line of each test case contains five integers a_1, a_2, a_3, a_4, a_5 (0 ≤ a_i ≤ 10^8), where a_i is the number of items of the i-th category Monocarp has to throw out (i = 1 is paper waste, i = 2 is plastic waste, i = 3 is general waste, i = 4 is partially-paper waste, i = 5 is partially-plastic waste).

Output

For each test case, print either YES if it is possible to fit all items into containers, or NO otherwise. You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).

Example

Input


7
1 2 3
1 2 3 0 0
2 2 3
1 2 3 1 0
2 2 3
1 2 3 0 1
1 2 5
1 2 3 1 1
0 0 0
0 0 0 0 0
0 0 4
1 0 0 0 0
13 37 42
0 0 0 40 47


Output


YES
YES
NO
YES
YES
NO
YES

Note

Explanations for the example test cases:

  1. Monocarp can put 1 item of paper waste into the first container, 2 items of plastic waste into the second container, and 3 items of general waste into the third container; 
  2. Monocarp can put 1 item of paper waste and 1 item of partially-paper waste into the first container, 2 items of plastic waste into the second container, and 3 items of general waste into the third container; 
  3. there is no answer since either the second container should contain 3 items, or the third container should contain 4 items; 
  4. Monocarp can put 1 item of paper waste into the first container, 2 items of plastic waste into the second container, and 3 items of general waste, 1 item of partially-paper waste and 1 item of partially-plastic waste into the third container; 
  5. there is no waste at all, so all containers can be left empty; 
  6. there's no answer since it's impossible to put a paper item into the third container; 
  7. Monocarp can put 10 items of partially-paper waste into the first container, 37 items of partially-plastic waste into the second container, and 30 items of partially-paper waste and 10 items of partially-plastic waste into the third container. 

Solution (in PYTHON3):

Solution:
#Code by Sounak, IIESTS
#------------------------------warmup----------------------------

import os
import sys
import math
from io import BytesIO, IOBase
from fractions import Fraction
import collections
from itertools import permutations
from collections import defaultdict
from collections import deque
import threading

#sys.setrecursionlimit(300000)
#threading.stack_size(10**8)

BUFSIZE = 8192
 
 
class FastIO(IOBase):
    newlines = 0
    
    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None
     
    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()
 
    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()
 
    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)
 
 
class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")
  
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")

#-------------------------------------------------------------------------
#mod = 9223372036854775807  
class SegmentTree:
    def __init__(self, data, default=10**10, func=lambda a, b: min(a,b)):
        """initialize the segment tree with data"""
        self._default = default
        self._func = func
        self._len = len(data)
        self._size = _size = 1 << (self._len - 1).bit_length()
 
        self.data = [default] * (2 * _size)
        self.data[_size:_size + self._len] = data
        for i in reversed(range(_size)):
            self.data[i] = func(self.data[i + i], self.data[i + i + 1])
 
    def __delitem__(self, idx):
        self[idx] = self._default
 
    def __getitem__(self, idx):
        return self.data[idx + self._size]
 
    def __setitem__(self, idx, value):
        idx += self._size
        self.data[idx] = value
        idx >>= 1
        while idx:
            self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
            idx >>= 1
 
    def __len__(self):
        return self._len
 
    def query(self, start, stop):
        if start == stop:
            return self.__getitem__(start)
        stop += 1
        start += self._size
        stop += self._size
 
        res = self._default
        while start < stop:
            if start & 1:
                res = self._func(res, self.data[start])
                start += 1
            if stop & 1:
                stop -= 1
                res = self._func(res, self.data[stop])
            start >>= 1
            stop >>= 1
        return res
 
    def __repr__(self):
        return "SegmentTree({0})".format(self.data)
    
MOD=10**9+7
class Factorial:
    def __init__(self, MOD):
        self.MOD = MOD
        self.factorials = [1, 1]
        self.invModulos = [0, 1]
        self.invFactorial_ = [1, 1]
 
    def calc(self, n):
        if n <= -1:
            print("Invalid argument to calculate n!")
            print("n must be non-negative value. But the argument was " + str(n))
            exit()
        if n < len(self.factorials):
            return self.factorials[n]
        nextArr = [0] * (n + 1 - len(self.factorials))
        initialI = len(self.factorials)
        prev = self.factorials[-1]
        m = self.MOD
        for i in range(initialI, n + 1):
            prev = nextArr[i - initialI] = prev * i % m
        self.factorials += nextArr
        return self.factorials[n]
 
    def inv(self, n):
        if n <= -1:
            print("Invalid argument to calculate n^(-1)")
            print("n must be non-negative value. But the argument was " + str(n))
            exit()
        p = self.MOD
        pi = n % p
        if pi < len(self.invModulos):
            return self.invModulos[pi]
        nextArr = [0] * (n + 1 - len(self.invModulos))
        initialI = len(self.invModulos)
        for i in range(initialI, min(p, n + 1)):
            next = -self.invModulos[p % i] * (p // i) % p
            self.invModulos.append(next)
        return self.invModulos[pi]
 
    def invFactorial(self, n):
        if n <= -1:
            print("Invalid argument to calculate (n^(-1))!")
            print("n must be non-negative value. But the argument was " + str(n))
            exit()
        if n < len(self.invFactorial_):
            return self.invFactorial_[n]
        self.inv(n)  # To make sure already calculated n^-1
        nextArr = [0] * (n + 1 - len(self.invFactorial_))
        initialI = len(self.invFactorial_)
        prev = self.invFactorial_[-1]
        p = self.MOD
        for i in range(initialI, n + 1):
            prev = nextArr[i - initialI] = (prev * self.invModulos[i % p]) % p
        self.invFactorial_ += nextArr
        return self.invFactorial_[n]
 
 
class Combination:
    def __init__(self, MOD):
        self.MOD = MOD
        self.factorial = Factorial(MOD)
 
    def ncr(self, n, k):
        if k < 0 or n < k:
            return 0
        k = min(k, n - k)
        f = self.factorial
        return f.calc(n) * f.invFactorial(max(n - k, k)) * f.invFactorial(min(k, n - k)) % self.MOD
mod=10**9+7
omod=998244353
#-------------------------------------------------------------------------
for _ in range (int(input())):
    c1,c2,c3=map(int,input().split())
    a=list(map(int,input().split()))
    ch=1
    if a[0]>c1 or a[1]>c2 or a[2]>c3:
        ch=0
        print("No")
        continue
    c1-=a[0]
    c2-=a[1]
    c3-=a[2]
    a[3]-=min(a[3],c1)
    a[4]-=min(a[4],c2)
    if a[3]+a[4]>c3:
        print("NO")
        continue
    print("YES")