You are given a sequence s consisting of n digits from 1 to 9.

You have to divide it into at least two segments (segment — is a consecutive sequence of elements) (in other words, you have to place separators between some digits of the sequence) in such a way that each element belongs to exactly one segment and if the resulting division will be represented as an integer numbers sequence then each next element of this sequence will be strictly greater than the previous one.

More formally: if the resulting division of the sequence is t_1, t_2, ..., t_k, where k is the number of element in a division, then for each i from 1 to k-1 the condition t_{i} < t_{i + 1} (using numerical comparing, it means that the integer representations of strings are compared) should be satisfied.

For example, if s=654 then you can divide it into parts [6, 54] and it will be suitable division. But if you will divide it into parts [65, 4] then it will be bad division because 65 > 4. If s=123 then you can divide it into parts [1, 23], [1, 2, 3] but not into parts [12, 3].

Your task is to find any suitable division for each of the q independent queries.

Input

The first line of the input contains one integer q (1 ≤ q ≤ 300) — the number of queries.

The first line of the i-th query contains one integer number n_i (2 ≤ n_i ≤ 300) — the number of digits in the i-th query.

The second line of the i-th query contains one string s_i of length n_i consisting only of digits from 1 to 9.

Output

If the sequence of digits in the i-th query cannot be divided into at least two parts in a way described in the problem statement, print the single line "NO" for this query.

Otherwise in the first line of the answer to this query print "YES", on the second line print k_i — the number of parts in your division of the i-th query sequence and in the third line print k_i strings t_{i, 1}, t_{i, 2}, ..., t_{i, k_i} — your division. Parts should be printed in order of the initial string digits. It means that if you write the parts one after another without changing their order then you'll get the string s_i.

See examples for better understanding.

Example

Input


4
6
654321
4
1337
2
33
4
2122


Output


YES
3
6 54 321
YES
3
1 3 37
NO
YES
2
21 22

import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;


 
public class TryC
{
    
    static class InputReader {
 
		private final InputStream stream;
		private final byte[] buf = new byte[8192];
		private int curChar, snumChars;
 
		public InputReader(InputStream st) {
			this.stream = st;
		}
 
		public int read() {
			if (snumChars == -1)
				throw new InputMismatchException();
			if (curChar >= snumChars) {
				curChar = 0;
				try {
					snumChars = stream.read(buf);
				} catch (IOException e) {
					throw new InputMismatchException();
				}
				if (snumChars <= 0)
					return -1;
			}
			return buf[curChar++];
		}
 
		public int ni() {
			int c = read();
			while (isSpaceChar(c)) {
				c = read();
			}
			int sgn = 1;
			if (c == '-') {
				sgn = -1;
				c = read();
			}
			int res = 0;
			do {
				res *= 10;
				res += c - '0';
				c = read();
			} while (!isSpaceChar(c));
			return res * sgn;
		}
 
		public long nl() {
			int c = read();
			while (isSpaceChar(c)) {
				c = read();
			}
			int sgn = 1;
			if (c == '-') {
				sgn = -1;
				c = read();
			}
			long res = 0;
			do {
				res *= 10;
				res += c - '0';
				c = read();
			} while (!isSpaceChar(c));
			return res * sgn;
		}
 
		public int[] nia(int n) {
			int a[] = new int[n];
			for (int i = 0; i < n; i++) {
				a[i] = ni();
			}
			return a;
		}
 
		public String rs() {
			int c = read();
			while (isSpaceChar(c)) {
				c = read();
			}
			StringBuilder res = new StringBuilder();
			do {
				res.appendCodePoint(c);
				c = read();
			} while (!isSpaceChar(c));
			return res.toString();
		}
 
		public String nextLine() {
			int c = read();
			while (isSpaceChar(c))
				c = read();
			StringBuilder res = new StringBuilder();
			do {
				res.appendCodePoint(c);
				c = read();
			} while (!isEndOfLine(c));
			return res.toString();
		}
 
		public boolean isSpaceChar(int c) {
			return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
		}
 
		private boolean isEndOfLine(int c) {
			return c == '\n' || c == '\r' || c == -1;
		}
 
	}
	

	
	
	
    static long mod=1000000007;
    static BigInteger bigInteger = new BigInteger("1000000007");
    static int n = (int)1e6;
    static boolean[] prime;
    static ArrayList<Integer> as;
     static ArrayList<Integer> []as1;
    static void sieve() {
			Arrays.fill(prime	, true);
			prime[0] = prime[1] = false;
			for(int i = 2 ; i * i <= n ; ++i) {
				if(prime[i]) {
					for(int k = i * i; k< n ; k+=i) {
						prime[k] = false;
					}
				}
			}
			
		}
		static PrintWriter w = new PrintWriter(System.out);
                static int p = 0;
 static char [][]sol;
    public static void main(String[] args)
    {
        
        InputReader sc = new InputReader(System.in);
        //PrintWriter w = new PrintWriter(System.out);
    
        
        prime  = new boolean[n + 1];
		sieve();
		prime[1] = false;
		
		/*
		as = new ArrayList<>();
		for(int i=2;i<=1000000;i++)
		{
		    if(prime[i])
		    as.add(i);
		}
		*/
		
       /* 
		
	long a = sc.nl();
	 
            BigInteger ans = new BigInteger("1");
            for (long i = 1; i < Math.sqrt(a); i++) {
                if (a % i == 0) {
                    if (a / i == i) {
                        ans = ans.multiply(BigInteger.valueOf(phi(i)));
                        
                    } else {
                        ans = ans.multiply(BigInteger.valueOf(phi(i)));
                        ans = ans.multiply(BigInteger.valueOf(phi(a / i)));
 
                        
                    }
 
 
                }
            }
            w.println(ans.mod(bigInteger));
            */
          
         // MergeSort ob = new MergeSort();
       // ob.sort(arr, 0, arr.length-1);
       
    // Student []st = new Student[x];    
   // st[i] = new Student(i,d[i]);
     //Arrays.sort(st,(p,q)->p.diff-q.diff);
     
  int q = sc.ni();
  while(q-- > 0)
  {
      int x = sc.ni();
      String s = sc.rs();
      if(x > 2)
      {
          w.println("YES");
          w.println("2");
          w.println(s.substring(0,1) + " " + s.substring(1,x));
          
      }
      else
      {
          if(ip(s.substring(0,1)) >= ip(s.substring(1,2)))
              w.println("NO");
          else
          {
              w.println("YES");
              w.println("2");
              w.println(s.charAt(0) + " " + s.charAt(1));
          }
          
      }
      
       
  }
  
   w.close();
        
  
    
    }
    static long log2(long value) {
    return Long.SIZE-Long.numberOfLeadingZeros(value);
}
   
     static class Student
    {
        
        int first;
        int sec;
        int index;
        Student(int first,int sec,int index)
        {
            this.first = first;
            this.sec = sec;
            this.index = index;             
            
        }
        
    }
     
     public static long modMultiply(long one, long two) {
 
		return (one % mod * two % mod) % mod;
	}
    static long fx(int m)
    {
        long re = 0;
        for(int i=1;i<=m;i++)
        {
            re += (long) (i / gcd(i,m));
        }
        return re;
    }
     static int gcd(int a, int b)
    {
	 if (a == 0)
	  return b;
				         
	return gcd(b%a, a);
    }
    
   
    
     static long phi(long nx)
    {
        // Initialize result as n
        double result = nx; 
        
     
        // Consider all prime factors of n and for 
        // every prime factor p, multiply result
        // with (1 - 1/p)
        for (int p = 0; as.get(p) * as.get(p) <= nx; ++p)
        {
            // Check if p is a prime factor.
            if (nx % as.get(p) == 0)
            {
                // If yes, then update n and result
                while (nx % as.get(p) == 0)
                    nx /= as.get(p);
                result *= (1.0 - (1.0 / (double) as.get(p)));
            }
        }
     
        // If n has a prime factor greater than sqrt(n)
        // (There can be at-most one such prime factor)
        if (nx > 1)
            result *= (1.0 - (1.0 / (double) nx));
     
        return (long)result;
    
        
        //return(phi((long)result,k-1));
        
    }
    
    public static void primeFactors(int n,int x)
    {
        as = new ArrayList<>();
       
        // Print the number of 2s that divide n
        while (n%2==0)
        {
            
        
            as.add(2);
            //System.out.print(2 + " ");
            n /= 2;
            
        }
 
        // n must be odd at this point.  So we can
        // skip one element (Note i = i +2)
        for (int i = 3; i <= Math.sqrt(n); i+= 2)
        {
            // While i divides n, print i and divide n
            while (n%i == 0)
            {
            
               // System.out.print(i + " ");
                
                as.add(i);
                n /= i;
             
                
            }
            
        }
 
        // This condition is to handle the case whien
        // n is a prime number greater than 2
        if (n >= 2)
        {
          
            as.add(n);
        }
           
    }
    static int digitsum(int x)
    
    {    
 
        int sum = 0;
        while(x > 0)
        {
            int temp = x % 10;
            sum += temp;
            x /= 10;
        }
        return sum;
        
    }
    static int countDivisors(int n)
    {
        int cnt = 0;
        for (int i = 1; i*i <=n; i++)
        {
             if (n % i == 0 && i<=1000000)
             {
                // If divisors are equal,
                // count only one
                if (n / i == i)
                   cnt++;
      
                else // Otherwise count both
                    cnt = cnt + 2;
             }
         }
        return cnt;
    }
   
    static boolean isprime(long n)
    {
    
    if(n == 2)
        return true;
    if(n == 3)
        return true;
    if(n % 2 == 0)
        return false;
    if(n % 3 == 0)
        return false;
 
    long i = 5;
   long w = 2;
 
    while(i * i <= n)
    {
        if(n % i == 0)
            return false;
    
 
        i += w;
        w = 6 - w;
    }
 
    return true;
    }
    
   
 
 
    
    
    static boolean binarysearch(int []arr,int p,int n)
    {
        //ArrayList<Integer> as = new ArrayList<>();
        //as.addAll(0,at);
        //Collections.sort(as);
        boolean re = false;
        int st = 0;
        int end = n-1;
        
        
        
        while(st <= end)
        { 
            int mid = (st+end)/2;
        
            if(p > arr[mid])
            {
                st = mid+1;
                
            }
            else if(p < arr[mid])
            {
                end = mid-1;
                
            }
            else if(p == arr[mid])
            {
            
            re = true;
            break;
            }
           
            
            
        }
         return re;
        
    }
    
   
    
    /* Java program for Merge Sort */
static class MergeSort
{
    // Merges two subarrays of arr[].
    // First subarray is arr[l..m]
    // Second subarray is arr[m+1..r]
    void merge(int arr[], int l, int m, int r)
    {
        // Find sizes of two subarrays to be merged
        int n1 = m - l + 1;
        int n2 = r - m;
 
        /* Create temp arrays */
        int L[] = new int [n1];
        int R[] = new int [n2];
 
        /*Copy data to temp arrays*/
        for (int i=0; i<n1; ++i)
            L[i] = arr[l + i];
        for (int j=0; j<n2; ++j)
            R[j] = arr[m + 1+ j];
 
 
        /* Merge the temp arrays */
 
        // Initial indexes of first and second subarrays
        int i = 0, j = 0;
 
        // Initial index of merged subarry array
        int k = l;
        while (i < n1 && j < n2)
        {
            if (L[i] <= R[j])
            {
                arr[k] = L[i];
                i++;
            }
            else
            {
                arr[k] = R[j];
                j++;
            }
            k++;
        }
 
        /* Copy remaining elements of L[] if any */
        while (i < n1)
        {
            arr[k] = L[i];
            i++;
            k++;
        }
 
        /* Copy remaining elements of R[] if any */
        while (j < n2)
        {
            arr[k] = R[j];
            j++;
            k++;
        }
    }
 
    // Main function that sorts arr[l..r] using
    // merge()
    void sort(int arr[], int l, int r)
    {
        if (l < r)
        {
            // Find the middle point
            int m = (l+r)/2;
 
            // Sort first and second halves
            sort(arr, l, m);
            sort(arr , m+1, r);
 
            // Merge the sorted halves
            merge(arr, l, m, r);
        }
    }
 
    /* A utility function to print array of size n */
    
 
   
}
    

    
    
    
    public static int ip(String s){
		return Integer.parseInt(s);
	}
	 public static String ips(int s){
		return Integer.toString(s);
	}
	
         // Java program to print all permutations using
// Heap's algorithm
static class HeapAlgo
{
   //static ArrayList<Integer> []ad;
	//Prints the array
	void printArr(int a[], int n)
	{
	    
	    
		for (int i=0; i<n; i++)
		{
			System.out.print(a[i] + " ");
		as1[p].add(a[i]);
		}
                p++;
                
		//System.out.println();
	}

	//Generating permutation using Heap Algorithm
	void heapPermutation(int a[], int size, int n)
	{
		// if size becomes 1 then prints the obtained
		// permutation
		
		if (size == 1)
		{
			printArr(a,n);
			
		}

		for (int i=0; i<size; i++)
		{
			heapPermutation(a, size-1, n);
                        p++;

			// if size is odd, swap first and last
			// element
			if (size % 2 == 1)
			{
				int temp = a[0];
				a[0] = a[size-1];
				a[size-1] = temp;
			}

			// If size is even, swap ith and last
			// element
			else
			{
				int temp = a[i];
				a[i] = a[size-1];
				a[size-1] = temp;
			}
		}
	}
        public static int lowerBound(int[] array, int length, int value) {
        int low = 0;
        int high = length-1;
        while (low < high) {
            final int mid = (low + high) / 2;
            if (value <= array[mid]) {
                high = mid;
            } else {
                low = mid + 1;
            }
        }
        return low;
  }
  public static int upperBound(int[] array, int length, int value) {
        int low = 0;
        int high = length-1;
        while (low < high) {
            final int mid = (low + high) / 2;
            if (value >= array[mid]) {
                low = mid + 1;
            } else {
                high = mid;
            }
        }
        return low;
    }

	// Driver code
	
		
	
}

// This code has been contributed by Amit Khandelwal.

    
					 
}