human
Can you provide a Python program to print the answer in one line given the following:
Anna is a girl so brave that she is loved by everyone in the city and citizens love her cookies. She is planning to hold a party with cookies. Now she has a vanilla cookies and b chocolate cookies for the party.

She invited n guests of the first type and m guests of the second type to the party. They will come to the party in some order. After coming to the party, each guest will choose the type of cookie (vanilla or chocolate) to eat. There is a difference in the way how they choose that type:

If there are v vanilla cookies and c chocolate cookies at the moment, when the guest comes, then

  * if the guest of the first type: if v>c the guest selects a vanilla cookie. Otherwise, the guest selects a chocolate cookie. 
  * if the guest of the second type: if v>c the guest selects a chocolate cookie. Otherwise, the guest selects a vanilla cookie. 


After that:

  * If there is at least one cookie of the selected type, the guest eats one. 
  * Otherwise (there are no cookies of the selected type), the guest gets angry and returns to home. 


Anna wants to know if there exists some order of guests, such that no one guest gets angry. Your task is to answer her question.

Input

The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases. Next t lines contain descriptions of test cases.

For each test case, the only line contains four integers a, b, n, m (0 ≤ a,b,n,m ≤ 10^{18}, n+m ≠ 0).

Output

For each test case, print the answer in one line. If there exists at least one valid order, print "Yes". Otherwise, print "No".

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

Example

Input

6
2 2 1 2
0 100 0 1
12 13 25 1
27 83 14 25
0 0 1 0
1000000000000000000 1000000000000000000 1000000000000000000 1000000000000000000

Output

Yes
No
No
Yes
No
Yes

Note

In the first test case, let's consider the order \{1, 2, 2\} of types of guests. Then:

  * The first guest eats a chocolate cookie. After that, there are 2 vanilla cookies and 1...
The above is tricky. Write me a correct solution in PYTHON3.
assistant
#Here is Python program to print the answer in one line.

# import math
# from collections import Counter
#from collections import defaultdict
#from collections import deque
# from bisect import bisect_left as bl, bisect_right as br
# from itertools import accumulate,permutations as perm,combinations as comb
# import heapq
# from functools import reduce
# from fractions import Fraction

def main():
    t = Iint()
    for _ in range(t):
        v, c, t1, t2 = Imap()  # van,choc,guest1,guest2
        if v+c < t1+t2:
            print("No")
        elif (v > c and c >= t2) or (v <= c and v >= t2):
            print("Yes")
        else:
            print("No")

# IO region
FIO = 1

def I(): return input()

def Iint(): return int(input())

def Ilist(): return list(map(int, input().split()))   # int list

def Imap(): return map(int, input().split())   # int map

def Plist(li, s=' '): print(s.join(map(str, li)))   # non string list

# /IO region
en = enumerate

if not FIO:
    if __name__ == '__main__':
        main()
        exit()
else:
    from io import BytesIO, IOBase
    import sys
    import os
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...