You are given the array a consisting of n positive (greater than zero) integers.

In one move, you can choose two indices i and j (i ≠ j) such that the absolute difference between a_i and a_j is no more than one (|a_i - a_j| ≤ 1) and remove the smallest of these two elements. If two elements are equal, you can remove any of them (but exactly one).

Your task is to find if it is possible to obtain the array consisting of only one element using several (possibly, zero) such moves or not.

You have to answer t independent test cases.

Input

The first line of the input contains one integer t (1 ≤ t ≤ 1000) — the number of test cases. Then t test cases follow.

The first line of the test case contains one integer n (1 ≤ n ≤ 50) — the length of a. The second line of the test case contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 100), where a_i is the i-th element of a.

Output

For each test case, print the answer: "YES" if it is possible to obtain the array consisting of only one element using several (possibly, zero) moves described in the problem statement, or "NO" otherwise.

Example

Input


5
3
1 2 2
4
5 5 5 5
3
1 2 4
4
1 3 4 4
1
100


Output


YES
YES
NO
NO
YES

Note

In the first test case of the example, we can perform the following sequence of moves:

  * choose i=1 and j=3 and remove a_i (so a becomes [2; 2]); 
  * choose i=1 and j=2 and remove a_j (so a becomes [2]). 



In the second test case of the example, we can choose any possible i and j any move and it doesn't matter which element we remove.

In the third test case of the example, there is no way to get rid of 2 and 4.

import java.io.*;
import java.io.BufferedReader; 
import java.io.IOException; 
import java.io.InputStreamReader;
import java.math.BigInteger;
import java.util.*;


/**
 * @author Utkarsh
 *
 */
public  class FibinacciSeries 
{

	   
	   static class FastReader 
	    { 
	        BufferedReader br; 
	        StringTokenizer st; 
	       
	  
	        public FastReader() 
	        { 
	            br = new BufferedReader(new
	                     InputStreamReader(System.in)); 
	        } 
	  
	        String next() 
	        { 
	            while (st == null || !st.hasMoreTokens()) 
	            { 
	                try
	                { 
	                    st = new StringTokenizer(br.readLine()); 
	                } 
	                catch (IOException  e) 
	                { 
	                    e.printStackTrace(); 
	                } 
	            } 
	            return st.nextToken();
	            
	        } 
	  
	        int nextInt() 
	        { 
	            return Integer.parseInt(next()); 
	        } 
	  
	        long nextLong() 
	        { 
	            return Long.parseLong(next()); 
	        } 
	  
	        double nextDouble() 
	        { 
	            return Double.parseDouble(next()); 
	        } 
	  
	        String nextLine() 
	        { 
	            String str = ""; 
	            try
	            { 
	                str = br.readLine(); 
	            } 
	            catch (IOException e) 
	            { 
	                e.printStackTrace(); 
	            } 
	            return str; 
	        }

			public Character charAt(int i) {
				// TODO Auto-generated method stub
				return null;
			}

			public BigInteger nextBigInteger() {
				// TODO Auto-generated method stub
				return null;
			} 
	    } 

  

	  // Complete the hurdleRace function below.
//    public static void solve() {
//    	FastReader s=new FastReader();
	    int binarySearch(int arr[], int l, int r, int x) 
	    { 
	        if (r >= l) { 
	            int mid = l + (r - l) / 2; 
	  
	            // If the element is present at the 
	            // middle itself 
	            if (arr[mid] == x) 
	                return mid; 
	  
	            // If element is smaller than mid, then 
	            // it can only be present in left subarray 
	            if (arr[mid] > x) 
	                return binarySearch(arr, l, mid - 1, x); 
	  
	            // Else the element can only be present 
	            // in right subarray 
	            return binarySearch(arr, mid + 1, r, x); 
	        } 
	  
	        // We reach here when element is not present 
	        // in array 
	        return -1; 
	    } 
	  
	    static // Driver method to test above 

	// Java implementation of iterative Binary Search 

	    // Returns index of x if it is present in arr[], 
	    // else return -1 
	    int lowerBound(int[] a,int n,int key){
	        int s =0,e=n-1;
	        int ans = -1;

	        while(s<=e){
	            int mid = (s+e)/2;

	            if(a[mid]==key){
	                ans = mid;
	                e = mid - 1;
	            }
	            else if(a[mid]>key){
	                e = mid - 1;
	            }
	            else{
	                s = mid + 1;
	            }
	        }

	        return ans;
	    }
	    static int count(String se,char c,int l,int r) {
			if(l==r) {
				if(se.charAt(l)==c)return 0;
				else return 1;
			}
			int mid = (l+r)/2;
			int c1=0,c2=0;
			for(int i=l;i<=mid;i++) {
				if(se.charAt(i)!=c)c1++;
			}
			for(int i=r;i>mid;i--) {
				if(se.charAt(i)!=c)c2++;
			}
			int y = count(se,(char)((int)c+1),mid+1,r);
			int x = count(se,(char)((int)c+1),l,mid);
	    	return Math.min(c1+y, c2+x);
	    	
	    }
	    public static double log2(long N) 
	    { 
	  
	        // calculate log2 N indirectly 
	        // using log() method 
	        double result = (double)(Math.log(N) / Math.log(2)); 
	  
	        return result; 
	    }
	    static boolean sign(long n) {
	    	return n>0;
	    }
	    public static int solve(long[] ar,int n) {
	    	long sum=0;
	    	for(int i=0;i<n;i++) {
	    		sum+=ar[i];
	    		if(sum<=0)return 0;
			}
	    	sum=0;
	    	for(int i=n-1;i>=0;i--) {
	    		sum+=ar[i];
	    		if(sum<=0)return 0;
	    	}
			return 1;
	    }
    public static void main(String[] args) throws IOException {
    	FastReader s=new FastReader();
    		int t = s.nextInt();
    		while(t-->0) {
    			int n = s.nextInt();
    			int max=0;
    			int[] ar = new int[n];
    			for(int i=0;i<n;i++) {
    				ar[i] =  s.nextInt();
    			}
    			Arrays.sort(ar);
    			for(int i=1;i<n;i++) {
    				max= Math.max(max, Math.abs(ar[i]-ar[i-1]));
    			}
    			if(max<=1)System.out.println("YES");
    			else System.out.println("NO");
    		}
    		
    	}
    	
    
    
}
//5
//4 8 2 6 2
//4 5 4 1 3



        
    

      
 