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 B {
  	public static void main(String args[]){
  		FScanner in = new FScanner();
  		PrintWriter out = new PrintWriter(System.out);
  		int t = in.nextInt();
  		while(t-->0) { 
  			int a=in.nextInt(),b=in.nextInt(),c=in.nextInt();
  			int m=in.nextInt();
  
  
  
  			long sum=a+b+c;
  			long max_p=sum;
  			if(a>0)
  				max_p--;
  			if(b>0)
  				max_p--;
  			if(c>0)
  				max_p--;
  			int max=Math.max(a,Math.max(b,c));
  			long sum_min=sum-max;
  
  			 if(m>max_p)
  			 	out.println("NO");
  			 else
  			 {
  			 	sum_min*=2;
  			 	sum_min++;
  
  			 	if(sum-(sum_min)>m)
  			 		out.println("NO");
  			 	else
  			 		out.println("YES");
  			 }
  		}
  		out.close();
  	}
  
  	static boolean checkprime(int n1)
  	{
  		if(n1%2==0||n1%3==0)
  			return false;
  		else 
  		{
  			for(int i=5;i*i<=n1;i+=6)
  			{
  				if(n1%i==0||n1%(i+2)==0)
  					return false;
  			}
  			return true;
  		}
  	}
  	
  	static class FScanner {
  		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
  		StringTokenizer sb = new StringTokenizer("");
  
  		String next(){
  			while(!sb.hasMoreTokens()){
  				try{
  					sb = new StringTokenizer(br.readLine());
  				} catch(IOException e){ }
  			}
  			return sb.nextToken();
  		}
  		String nextLine(){
  			try{ return br.readLine(); } 
  			catch(IOException e) { } return "";
  		}
  
  		int nextInt(){
  			return Integer.parseInt(next());
  		}
  
  		long nextLong() {
  			return Long.parseLong(next());
  		}
  
  		int[] readArray(int n) {
  			int a[] = new int[n];
  			for(int i=0;i<n;i++) a[i] = nextInt();
  			return a;
  		}
  
  		float nextFloat(){
  		return Float.parseFloat(next());
  		}
  		
  		double nextDouble(){
  		return Double.parseDouble(next());
  		}
  	}
  }
  