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.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author Pradyumn Agrawal (coderbond007)
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        FastScanner in = new FastScanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        FArrayPartition solver = new FArrayPartition();
        int testCount = Integer.parseInt(in.next());
        for (int i = 1; i <= testCount; i++)
            solver.solve(i, in, out);
        out.close();
    }

    static class FArrayPartition {
        public void solve(int testNumber, FastScanner in, PrintWriter out) {
            int n = in.nextInt();
            int[] a = in.nextIntArray(n);
            int[] preMax = a.clone();
            int[] sufMax = a.clone();
            for (int i = 1; i < n; ++i) {
                preMax[i] = Math.max(preMax[i - 1], preMax[i]);
            }
            for (int i = n - 2; i >= 0; --i) {
                sufMax[i] = Math.max(sufMax[i + 1], sufMax[i]);
            }
            int[] L = getLeft(a);
            int[] R = getRight(a);
            for (int i = 1; i < n - 1; ++i) {
                int left = L[i];
                int right = R[i];
                int ind1 = Arrays.binarySearch(preMax, Math.max(left, 0), i, a[i]);
                if (ind1 < 0) continue;
                int ind2 = binarySearch0(sufMax, i + 1, right > n ? n : (right + 1), a[i]);
                if (ind2 < 0) continue;
                out.println("YES");
                int x = ind1 + 1;
                int y = ind2 - x;
                int z = n - x - y;
                out.println(x + " " + y + " " + z);
                return;
            }
            out.println("NO");
        }

        private static int binarySearch0(int[] a, int fromIndex, int toIndex, int key) {
            int low = fromIndex;
            int high = toIndex - 1;

            while (low <= high) {
                int mid = (low + high) >>> 1;
                int midVal = a[mid];
                if (midVal < key)
                    high = mid - 1;
                else if (midVal > key)
                    low = mid + 1;
                else
                    return mid; // key found
            }
            return -(low + 1);  // key not found.
        }

        private int[] getRight(int[] a) {
            int n = a.length;
            int[] stack = new int[n];
            int[] indexes = new int[n];
            ArrayUtils.fill(indexes, n + 1);
            int ptr = 0;
            for (int i = 0; i < n; ++i) {
                while (ptr != 0 && a[stack[ptr - 1]] > a[i]) {
                    indexes[stack[--ptr]] = i;
                }
                stack[ptr++] = i;
            }
            return indexes;
        }

        private int[] getLeft(int[] a) {
            int n = a.length;
            int[] stack = new int[n];
            int[] indexes = new int[n];
            ArrayUtils.fill(indexes, -1);
            int ptr = 0;
            for (int i = n - 1; i >= 0; --i) {
                while (ptr != 0 && a[stack[ptr - 1]] > a[i]) {
                    indexes[stack[--ptr]] = i;
                }
                stack[ptr++] = i;
            }
            return indexes;
        }

    }

    static class ArrayUtils {
        public static void fill(int[] array, int value) {
            Arrays.fill(array, value);
        }

    }

    static class FastScanner {
        BufferedReader reader;
        StringTokenizer tokenizer;

        public FastScanner(InputStream inputStream) {
            reader = new BufferedReader(new InputStreamReader(inputStream), 32768);
            tokenizer = null;
        }

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

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

        public int[] nextIntArray(int n) {
            int[] arr = new int[n];
            for (int i = 0; i < n; i++) {
                arr[i] = nextInt();
            }
            return arr;
        }

    }
}