Problem: You are given an array a consisting of n integers.

Let min(l, r) be the minimum value among a_l, a_{l + 1}, …, a_r and max(l, r) be the maximum value among a_l, a_{l + 1}, …, a_r.

Your task is to choose three positive (greater than 0) integers x, y and z such that:

  * x + y + z = n; 
  * max(1, x) = min(x + 1, x + y) = max(x + y + 1, n). 



In other words, you have to split the array a into three consecutive non-empty parts that cover the whole array and the maximum in the first part equals the minimum in the second part and equals the maximum in the third part (or determine it is impossible to find such a partition).

Among all such triples (partitions), you can choose any.

You have to answer t independent test cases.

Input

The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.

The first line of the test case contains one integer n (3 ≤ n ≤ 2 ⋅ 10^5) — the length of a.

The second line of the test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9), where a_i is the i-th element of a.

It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).

Output

For each test case, print the answer: NO in the only line if there is no such partition of a that satisfies the conditions from the problem statement. Otherwise, print YES in the first line and three integers x, y and z (x + y + z = n) in the second line.

If there are several answers, you can print any.

Example

Input


6
11
1 2 3 3 3 4 4 3 4 2 1
8
2 9 1 7 3 9 4 1
9
2 1 4 2 4 3 3 1 2
7
4 2 1 1 4 1 4
5
1 1 1 1 1
7
4 3 4 3 3 3 4


Output


YES
6 1 4
NO
YES
2 5 2
YES
4 1 2
YES
1 1 3
YES
2 1 4

Solution (in JAVA):

Solution:
import java.util.*;
import java.io.*;

public class Main {
    public static void main(String args[]) {new Main().run();}

    FastReader in = new FastReader();
    PrintWriter out = new PrintWriter(System.out);
    void run(){
        for(int q=ni();q>0;q--){
            work();
        }
        out.flush();
    }
    long mod=1000000007;
    long gcd(long a,long b) {
        return a==0?b:gcd(b%a,a);
    }
    void work() {
        int n=ni();
        int[] A=nia(n);
        int[] L=new int[n];
        int[] R=new int[n];
        int[] idxL=new int[n];
        int[] idxR=new int[n];
        int[] recl=new int[n];
        int[] recr=new int[n];
        for(int i=0,max=0;i<n;i++){
            max=Math.max(max,A[i]);
            L[i]=max;
        }
        for(int i=n-1,max=0;i>=0;i--){
            max=Math.max(max,A[i]);
            R[i]=max;
        }
        TreeMap<Integer,Integer> map=new TreeMap<>();
        for(int i=0;i<n;i++){
            int v=A[i];
            Integer d = map.higherKey(v);
            if(d==null){
                idxL[i]=i;
                if(!map.containsKey(v))map.put(v,i);
            }else{
                idxL[i]=map.get(d);
            }
        }
        map=new TreeMap<Integer, Integer>();
        for(int i=n-1;i>=0;i--){
            int v=A[i];
            Integer d = map.higherKey(v);
            if(d==null){
                idxR[i]=i;
                if(!map.containsKey(v))map.put(v,i);
            }else{
                idxR[i]=map.get(d);
            }
        }
        recl[0]=-1;
        recr[n-1]=n;
        for(int i=1;i<n;i++){
            int idx=i-1;
            while(idx>=0&&A[idx]>=A[i]){
                idx=recl[idx];
            }
            recl[i]=idx;
        }
        for(int i=n-2;i>=0;i--){
            int idx=i+1;
            while(idx<n&&A[idx]>=A[i]){
                idx=recr[idx];
            }
            recr[i]=idx;
        }
        for(int i=1;i<n-1;i++){
            int idxl=idxL[i];
            int idxr=idxR[i];
            int l=recl[i];
            int r=recr[i];
            if(idxl>0&&idxr<n-1&&L[idxl-1]==A[i]&&R[idxr+1]==A[i]&&l<=idxl&&r>=idxr){
                out.println(("YES"));
                out.println(idxl+" "+(idxr-idxl+1)+" "+(n-idxr-1));
                return;
            }
        }
        out.println("NO");
    }

    //input
    @SuppressWarnings("unused")
    private ArrayList<Integer>[] ng(int n, int m) {
        ArrayList<Integer>[] graph=(ArrayList<Integer>[])new ArrayList[n];
        for(int i=0;i<n;i++) {
            graph[i]=new ArrayList<>();
        }
        for(int i=1;i<=m;i++) {
            int s=in.nextInt()-1,e=in.nextInt()-1;
            graph[s].add(e);
            graph[e].add(s);
        }
        return graph;
    }

    private ArrayList<long[]>[] ngw(int n, int m) {
        ArrayList<long[]>[] graph=(ArrayList<long[]>[])new ArrayList[n];
        for(int i=0;i<n;i++) {
            graph[i]=new ArrayList<>();
        }
        for(int i=1;i<=m;i++) {
            long s=in.nextLong()-1,e=in.nextLong()-1,w=in.nextLong();
            graph[(int)s].add(new long[] {e,w,i});
            graph[(int)e].add(new long[] {s,w});
        }
        return graph;
    }

    private int ni() {
        return in.nextInt();
    }

    private long nl() {
        return in.nextLong();
    }

    private String ns() {
        return in.next();
    }

    private long[] na(int n) {
        long[] A=new long[n];
        for(int i=0;i<n;i++) {
            A[i]=in.nextLong();
        }
        return A;
    }
    private int[] nia(int n) {
        int[] A=new int[n];
        for(int i=0;i<n;i++) {
            A[i]=in.nextInt();
        }
        return A;
    }
}

class FastReader
{
    BufferedReader br;
    StringTokenizer st;

    public FastReader()
    {
        br=new BufferedReader(new InputStreamReader(System.in));
    }


    public String next()
    {
        while(st==null || !st.hasMoreElements())//回车，空行情况
        {
            try {
                st = new StringTokenizer(br.readLine());
            } catch (IOException e) {
                e.printStackTrace();
            }
        }
        return st.nextToken();
    }

    public int nextInt()
    {
        return Integer.parseInt(next());
    }

    public long nextLong()
    {
        return Long.parseLong(next());
    }
}