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.math.*;

public class snack4 
{ 
    static class Reader 
    { 
        final private int BUFFER_SIZE = 1 << 16; 
        private DataInputStream din; 
        private byte[] buffer; 
        private int bufferPointer, bytesRead; 
  
        public Reader() 
        { 
            din = new DataInputStream(System.in); 
            buffer = new byte[BUFFER_SIZE]; 
            bufferPointer = bytesRead = 0; 
        } 
  
        public Reader(String file_name) throws IOException 
        { 
            din = new DataInputStream(new FileInputStream(file_name)); 
            buffer = new byte[BUFFER_SIZE]; 
            bufferPointer = bytesRead = 0; 
        } 
  
        public String readLine() throws IOException 
        { 
            byte[] buf = new byte[1000000]; // line length 
            int cnt = 0, c; 
            while ((c = read()) != -1) 
            { 
                if (c == '\n') 
                    break; 
                buf[cnt++] = (byte) c; 
            } 
            return new String(buf, 0, cnt); 
        } 
  
        public int nextInt() throws IOException 
        { 
            int ret = 0; 
            byte c = read(); 
            while (c <= ' ') 
                c = read(); 
            boolean neg = (c == '-'); 
            if (neg) 
                c = read(); 
            do
            { 
                ret = ret * 10 + c - '0'; 
            }  while ((c = read()) >= '0' && c <= '9'); 
  
            if (neg) 
                return -ret; 
            return ret; 
        } 
  
        public long nextLong() throws IOException 
        { 
            long ret = 0; 
            byte c = read(); 
            while (c <= ' ') 
                c = read(); 
            boolean neg = (c == '-'); 
            if (neg) 
                c = read(); 
            do { 
                ret = ret * 10 + c - '0'; 
            } 
            while ((c = read()) >= '0' && c <= '9'); 
            if (neg) 
                return -ret; 
            return ret; 
        } 
  
        public double nextDouble() throws IOException 
        { 
            double ret = 0, div = 1; 
            byte c = read(); 
            while (c <= ' ') 
                c = read(); 
            boolean neg = (c == '-'); 
            if (neg) 
                c = read(); 
  
            do { 
                ret = ret * 10 + c - '0'; 
            } 
            while ((c = read()) >= '0' && c <= '9'); 
  
            if (c == '.') 
            { 
                while ((c = read()) >= '0' && c <= '9') 
                { 
                    ret += (c - '0') / (div *= 10); 
                } 
            } 
  
            if (neg) 
                return -ret; 
            return ret; 
        } 
  
        private void fillBuffer() throws IOException 
        { 
            bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE); 
            if (bytesRead == -1) 
                buffer[0] = -1; 
        } 
  
        private byte read() throws IOException 
        { 
            if (bufferPointer == bytesRead) 
                fillBuffer(); 
            return buffer[bufferPointer++]; 
        } 
  
        public void close() throws IOException 
        { 
            if (din == null) 
                return; 
            din.close(); 
        } 
    } 
 static 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() 
    static 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); 
        } 
    }
    public static void main(String[] args) throws IOException 
    { 
      //  Reader s=new Reader(); 
        Scanner sa=new Scanner(System.in);
        int q=sa.nextInt();
        while(q--!=0){
            int n=sa.nextInt();
            String s=sa.next();
            int l=s.length();
            if(l==2){
                if(s.charAt(0)>=s.charAt(1)){
                    System.out.println("NO");
                }
                else{
                    System.out.println("YES");
                    System.out.println("2");
                    System.out.println(s.charAt(0)+" "+s.charAt(1));       
                }
            }
            else{
                    System.out.println("YES");
                    System.out.println("2");
                    System.out.println(s.charAt(0)+" "+s.substring(1));                
            }
        }
    }
    static boolean isPrime(int n) 
{ 
    // Corner cases 
    if (n <= 1) 
        return false; 
    if (n <= 3) 
        return true; 
  
    // This is checked so  
    // that we can skip 
    // middle five numbers 
    // in below loop 
    if (n % 2 == 0 ||  
        n % 3 == 0) 
        return false; 
  
    for (int i = 5; 
             i * i <= n; i = i + 6) 
        if (n % i == 0 || 
            n % (i + 2) == 0) 
            return false; 
  
    return true; 
} 
    static int gcd(int a, int b) 
    { 
      if (b == 0) 
        return a; 
      return gcd(b, a % b);  
    }  
    //void answer(int a,)
    // public static void sortbykey() 
    // { 
    //     // TreeMap to store values of HashMap 
    //     TreeMap<Integer, Integer> sorted = new TreeMap<>(); 
  
    //     // Copy all data from hashMap into TreeMap 
    //     sorted.putAll(count); 
  
    //     // Display the TreeMap which is naturally sorted 
    //     // for (Map.Entry<String, Integer> entry : sorted.entrySet())  
    //     //     // System.out.println("Key = " + entry.getKey() +  
    //     //                  ", Value = " + entry.getValue());         
    // } 
    static long fast_pow(long base,long n,long M)
    {
        if(n==0)
           return 1;
        if(n==1)
        return base;
        long halfn=fast_pow(base,n/2,M);
        if(n%2==0)
            return ( halfn * halfn ) % M;
        else
            return ( ( ( halfn * halfn ) % M ) * base ) % M;
    }
    static long findMMI_fermat(long n,int M)
    {
        return fast_pow(n,M-2,M);
        //gdzjhf
        //zrhzu
    }
} 
