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:
/*
    Java might be good for Competitive Programming but Python still rocks!!!!!!!!!


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   pppppppppppppppppppppppppppppppppppppppppppppp     pppppppppppppppppppppppppp
  ppppppppppppppppppppppppppppppppppppppppppppp      pppppppppppppppppppppppppppp
  pppppppppppppppppppppppppppppppp               pppppppppppppppppppppppppppppppp
  pppppppppppppppppppppppppppp     pppppppppppppppppppppppppppppppppppppppppppppp
  ppppppppppppppppppppppppppp    pppppppppppppppppppppppppppppppppppppppppppppppp
    pppppppppppppppppppppppp  pppppppppppppppppppppppppppppppppppppppppppppppppp
     ppppppppppppppppppppppp  ppppppppppppppppppppppppppppppppppppppppppppppppp
      pppppppppppppppppppppp  ppppppppppppppppppppppppppppppppppppppppppppppp
       ppppppppppppppppppppp  ppppppppppppppppppppppppppppppppppppppppppppp
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*/

/*
10
10
3 2 8 8 3 2 5 8 7 5
10
4 6 2 6 3 7 1 4 2 6
10
8 4 5 7 5 5 5 5 3 2
10
7 4 1 4 3 7 1 6 6 3
10
1 7 1 2 5 3 7 2 3 3
10
3 4 5 8 8 5 3 4 7 7
10
6 1 4 1 6 7 1 3 6 3
10
2 4 2 2 8 4 8 2 6 2
10
4 2 7 6 2 5 6 8 8 4
10
4 8 1 8 5 3 3 4 3 5


YES
3 1 6
NO
NO
NO
NO
NO
YES
4 2 4
YES
1 8 1
NO
NO
 */
import java.util.*;
import java.lang.*;
import java.io.*;
public final class ProblemF
{
    public static BufferedWriter out = new BufferedWriter(new OutputStreamWriter(System.out));
    public static int t, n, m, i, j, k, u, v, w, q, low, high, mid, inf = 1_000_000_008;
    public static int[] arr, st, maxi;
    public static void main(String[] args) throws IOException
    {
        FastScanner sc = new FastScanner();
        int left_max;
        t = sc.nextInt();
        boolean result;
        while(t-- > 0)
        {
            n = sc.nextInt();
            arr = new int[n + 1];
            maxi = new int[n + 1];
            st = new int[4 * n];
            HashMap<Integer, ArrayDeque<Integer>> index = new HashMap<>();
            for(i = 1; i <= n; i++)
                arr[i] = maxi[i] = sc.nextInt();
            index.put(maxi[n], new ArrayDeque<>());
            index.get(maxi[n]).addFirst(n);
            for(i = n - 1; i >= 1; i--)
            {
                maxi[i] = Math.max(maxi[i], maxi[i + 1]);
                if(!index.containsKey(maxi[i])) index.put(maxi[i], new ArrayDeque<>());
                index.get(maxi[i]).addFirst(i);
            }
            build(1, 1, n);
            left_max = arr[1];
            result = false;
            for(i = 2; i < n; i++)
            {
                if(index.containsKey(left_max))
                {
                    low = index.get(left_max).getFirst() - 1;
                    if (low < i) low = i;
                    high = index.get(left_max).getLast() - 1;
//                    System.out.println(i + " " + left_max + " " + low + " " + high);
                    int found = -1;
//                    System.out.println(low + " " + high);
                    while (low <= high)
                    {
                        mid = (low + high) >> 1;
                        q = query(1, 1, n, i, mid);
//                        System.out.println(mid + " " + q);
                        if (q == left_max)
                        {
                            found = mid;
                            break;
                        }
                        if (q > left_max) low = mid + 1;
                        else high = mid - 1;
                    }
                    if (found != -1)
                    {
                        out.write("YES\n" + (i - 1) + " " + (found - i + 1) + " " + (n - found) + "\n");
                        out.flush();
                        result = true;
                        break;
                    }
                }
                left_max = Math.max(left_max, arr[i]);
            }
            if(!result)
            {
                out.write("NO\n");
                out.flush();
            }
        }
        out.close();
    }

    public static int build(int node, int left, int right)
    {
        if(left == right)
            return st[node] = arr[left];
        int mid = (left + right) >> 1;
        return st[node] = Math.min(build(node * 2, left, mid), build(node * 2 + 1, mid + 1, right));
    }

    public static int query(int node, int left, int right, int start, int end)
    {
        if(left > right || start > end || start > right || left > end)
            return inf;
        if(start <= left && end >= right)
            return st[node];
        int mid = (left + right) >> 1;
        return Math.min(query(node * 2, left, mid, start, end), query(node * 2 + 1, mid + 1, right, start, end));
    }

    public static class Print
    {
        Print()
        {}

        void integer(int... arr) throws IOException
        {
            for(int value: arr)
            {
                out.write(value+" ");
                out.flush();
            }
            out.write("\n");
            out.flush();
        }

        void string(String... arr) throws IOException
        {
            for(String value: arr)
            {
                out.write(value+" ");
                out.flush();
            }
            out.write("\n");
            out.flush();
        }

        void character(char... arr) throws IOException
        {
            for(char value: arr)
            {
                out.write(value+" ");
                out.flush();
            }
            out.write("\n");
            out.flush();
        }

        void long_int(long... arr) throws IOException
        {
            for(long value: arr)
            {
                out.write(value+" ");
                out.flush();
            }
            out.write("\n");
            out.flush();
        }
    }

    static class FastScanner
    {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st=new StringTokenizer("");
        String next()
        {
            while (!st.hasMoreTokens())
                try
                {
                    st=new StringTokenizer(br.readLine());
                } catch (IOException e)
                {
                    e.printStackTrace();
                }
            return st.nextToken();
        }

        int nextInt()
        {
            return Integer.parseInt(next());
        }
        int[] readArray(int n)
        {
            int[] a=new int[n];
            for (int i=0; i<n; i++) a[i]=nextInt();
            return a;
        }
        long nextLong()
        {
            return Long.parseLong(next());
        }
    }
}