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: 
  
  import java.io.*;
  import java.util.*;
  
  public class Main2 {
  	public static void main(String[] args) throws Exception {
  		Scanner sc = new Scanner(System.in);
  		PrintWriter pw = new PrintWriter(System.out);
  		int t = sc.nextInt();
  		while (t-- > 0) {
  			int a = sc.nextInt();
  			int b = sc.nextInt();
  			int c = sc.nextInt();
  			int m = sc.nextInt();
  			boolean flag = true;
  			int max = Math.max(Math.max(a, b), c);
  			long s = a + b + c - max + 1;
  			if (!(m >= max - s)) {
  				flag = false;
  			}
  			if (a + b + c - 3 < m) {
  				flag = false;
  			}
  //			int dif =0;
  //			if(min == a) {
  //				dif = Math.abs(b-c);
  //			}else {
  //				if(min==b) {
  //					dif = Math.abs(a-c);
  //				}else {
  //					dif = Math.abs(a-b);
  //				}
  //			}
  			if (flag) {
  				pw.println("YES");
  			} else {
  				pw.println("NO");
  			}
  		}
  		pw.close();
  	}
  
  	// --------------------sTufF ----------------------
  	static class Scanner {
  		StringTokenizer st;
  		BufferedReader br;
  
  		public Scanner(InputStream s) {
  			br = new BufferedReader(new InputStreamReader(s));
  		}
  
  		public Scanner(FileReader r) {
  			br = new BufferedReader(r);
  		}
  
  		public String next() throws IOException {
  			while (st == null || !st.hasMoreTokens())
  				st = new StringTokenizer(br.readLine());
  			return st.nextToken();
  		}
  
  		public int nextInt() throws IOException {
  			return Integer.parseInt(next());
  		}
  
  		public long nextLong() throws IOException {
  			return Long.parseLong(next());
  		}
  
  		public String nextLine() throws IOException {
  			return br.readLine();
  		}
  
  		public double nextDouble() throws IOException {
  			String x = next();
  			StringBuilder sb = new StringBuilder("0");
  			double res = 0, f = 1;
  			boolean dec = false, neg = false;
  			int start = 0;
  			if (x.charAt(0) == '-') {
  				neg = true;
  				start++;
  			}
  			for (int i = start; i < x.length(); i++)
  				if (x.charAt(i) == '.') {
  					res = Long.parseLong(sb.toString());
  					sb = new StringBuilder("0");
  					dec = true;
  				} else {
  					sb.append(x.charAt(i));
  					if (dec)
  						f *= 10;
  				}
  			res += Long.parseLong(sb.toString()) / f;
  			return res * (neg ? -1 : 1);
  		}
  
  		public long[] nextlongArray(int n) throws IOException {
  			long[] a = new long[n];
  			for (int i = 0; i < n; i++)
  				a[i] = nextLong();
  			return a;
  		}
  
  		public Long[] nextLongArray(int n) throws IOException {
  			Long[] a = new Long[n];
  			for (int i = 0; i < n; i++)
  				a[i] = nextLong();
  			return a;
  		}
  
  		public int[] nextIntArray(int n) throws IOException {
  			int[] a = new int[n];
  			for (int i = 0; i < n; i++)
  				a[i] = nextInt();
  			return a;
  		}
  
  		public Integer[] nextIntegerArray(int n) throws IOException {
  			Integer[] a = new Integer[n];
  			for (int i = 0; i < n; i++)
  				a[i] = nextInt();
  			return a;
  		}
  
  		public boolean ready() throws IOException {
  			return br.ready();
  		}
  	}
  
  }
  