description: You are given four integer values a, b, c and m.
  
  Check if there exists a string that contains: 
  
    * a letters 'A'; 
    * b letters 'B'; 
    * c letters 'C'; 
    * no other letters; 
    * exactly m pairs of adjacent equal letters (exactly m such positions i that the i-th letter is equal to the (i+1)-th one). 
  
  Input
  
  The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of testcases.
  
  Each of the next t lines contains the description of the testcase — four integers a, b, c and m (1 ≤ a, b, c ≤ 10^8; 0 ≤ m ≤ 10^8).
  
  Output
  
  For each testcase print "YES" if there exists a string that satisfies all the requirements. Print "NO" if there are no such strings.
  
  You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES will all be recognized as positive answer).
  
  Example
  
  Input
  
  
  3
  2 2 1 0
  1 1 1 1
  1 2 3 2
  
  
  Output
  
  
  YES
  NO
  YES
  
  Note
  
  In the first testcase strings "ABCAB" or "BCABA" satisfy the requirements. There exist other possible strings.
  
  In the second testcase there's no way to put adjacent equal letters if there's no letter that appears at least twice.
  
  In the third testcase string "CABBCC" satisfies the requirements. There exist other possible strings.
solution: #!/usr/bin/env python3
  import sys
  import getpass  # not available on codechef
  import math, random
  import functools, itertools, collections, heapq, bisect
  from collections import Counter, defaultdict, deque
  input = sys.stdin.readline  # to read input quickly
  
  # available on Google, AtCoder Python3, not available on Codeforces
  # import numpy as np
  # import scipy
  
  M9 = 10**9 + 7  # 998244353
  yes, no = "YES", "NO"
  # d4 = [(1,0),(0,1),(-1,0),(0,-1)]
  # d8 = [(1,0),(1,1),(0,1),(-1,1),(-1,0),(-1,-1),(0,-1),(1,-1)]
  # d6 = [(2,0),(1,1),(-1,1),(-2,0),(-1,-1),(1,-1)]  # hexagonal layout
  MAXINT = sys.maxsize
  
  # if testing locally, print to terminal with a different color
  OFFLINE_TEST = getpass.getuser() == "hkmac"
  # OFFLINE_TEST = False  # codechef does not allow getpass
  def log(*args):
      if OFFLINE_TEST:
          print('\033[36m', *args, '\033[0m', file=sys.stderr)
  
  def solve(*args):
      # screen input
      if OFFLINE_TEST:
          log("----- solving ------")
          log(*args)
          log("----- ------- ------")
      return solve_(*args)
  
  def read_matrix(rows):
      return [list(map(int,input().split())) for _ in range(rows)]
  
  def read_strings(rows):
      return [input().strip() for _ in range(rows)]
  
  def minus_one(arr):
      return [x-1 for x in arr]
  
  def minus_one_matrix(mrr):
      return [[x-1 for x in row] for row in mrr]
  
  # ---------------------------- template ends here ----------------------------
  
  
  def solve_(a,b,c,m):
      # your solution here
  
      a,b,c = sorted([a,b,c])
  
      upper = max(0,a-1) + max(0,b-1) + max(0,c-1)
  
      log(upper)
  
      if m > upper:
          return no
  
      lower = (c - a - b - 1)
      lower = max(lower, 0)
  
      log(lower)
  
      if m < lower:
          return no
  
      return yes
  
  
  # for case_num in [0]:  # no loop over test case
  # for case_num in range(100):  # if the number of test cases is specified
  for case_num in range(int(input())):
  
      # read line as an integer
      # k = int(input())
  
      # read line as a string
      # srr = input().strip()
  
      # read one line and parse each word as a string
      # lst = input().split()
  
      # read one line and parse each word as an integer
      a,b,c,m = list(map(int,input().split()))
      # lst = list(map(int,input().split()))
      # lst = minus_one(lst)
  
      # read multiple rows
      # arr = read_strings(k)  # and return as a list of str
      # mrr = read_matrix(k)  # and return as a list of list of int
      # mrr = minus_one_matrix(mrr)
  
      res = solve(a,b,c,m)  # include input here
  
      # print length if applicable
      # print(len(res))
  
      # parse result
      # res = " ".join(str(x) for x in res)
      # res = "\n".join(str(x) for x in res)
      # res = "\n".join(" ".join(str(x) for x in row) for row in res)
  
      # print result
      # print("Case #{}: {}".format(case_num+1, res))   # Google and Facebook - case number required
  
      print(res)
  