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:
/**
 * author: derrick20
 * created: 11/25/20 5:23 PM
 */
import java.io.*;
import java.util.*;
import static java.lang.Math.*;

public class ArrayPartitionNice {
    public static void main(String[] args) {
        FastScanner sc = new FastScanner();
        PrintWriter out = new PrintWriter(System.out);

        int T = sc.nextInt();
        cases: while (T-->0) {
            int N = sc.nextInt();
            int[] a = new int[N + 1];
            int[] prefix = new int[N + 1];
            int[] suffix = new int[N + 2];
            for (int i = 1; i <= N; i++) {
                a[i] = sc.nextInt();
                prefix[i] = max(prefix[i - 1], a[i]);
            }
            for (int i = N; i >= 1; i--) {
                suffix[i] = max(suffix[i + 1], a[i]);
            }
            int bound = prefix[N];
            int middle = bound;
            int curr = middle;
            boolean seen = false;
            int l = -1, r = -1;
            for (int i = 1; i <= N; i++) {
                if (seen) {
                    curr = min(curr, a[i]);
                }
                if (a[i] == bound) {
                    if (!seen) {
                        l = i;
                        seen = true;
                    }
                    if (prefix[i - 1] == a[i] && a[i] == suffix[i + 1]) {
                        int x = i - 1;
                        int y = 1;
                        int z = N - i;
                        out.println("YES");
                        out.println(x + " " + y + " " + z);
                        continue cases;
                    }
                    middle = curr;
                    r = i;
                }
            }
            while (2 <= l && r <= N - 1) {
//                System.out.println(l + " " + r);
                if (prefix[l - 1] == middle && middle == suffix[r + 1]) {
                    int x = l - 1;
                    int y = r - l + 1;
                    int z = N - r;
                    out.println("YES");
                    out.println(x + " " + y + " " + z);
                    continue cases;
                }
                /*
                Actually, we can reduce all the logic into just making the minimum
                of all three maximized. This is because all three of our groups will
                monotonically decrease as we expand the middle, no matter what we do.
                So, this is the only strategy that ensures we don't skip any potential
                solutions where u = v = w = k, for some k. By exchange, if we choose
                anything else, we only increase the possibility of skipping over a k.
                 */
                TreeSet<Integer> left = new TreeSet<>(Arrays.asList(prefix[l - 2], min(middle, a[l - 1]), suffix[r + 1]));
                TreeSet<Integer> right = new TreeSet<>(Arrays.asList(prefix[l - 1], min(middle, a[r + 1]), suffix[r + 2]));
                if (left.first() >= right.first()) {
                    middle = min(middle, a[l - 1]);
                    l--;
                } else {
                    middle = min(middle, a[r + 1]);
                    r++;
                }
            }
            out.println("NO");
        }
        out.close();
    }
/*
1
6
1 2 3 2 1 2
 */
    static class FastScanner {
        private int BS = 1 << 16;
        private char NC = (char) 0;
        private byte[] buf = new byte[BS];
        private int bId = 0, size = 0;
        private char c = NC;
        private double cnt = 1;
        private BufferedInputStream in;

        public FastScanner() {
            in = new BufferedInputStream(System.in, BS);
        }

        public FastScanner(String s) {
            try {
                in = new BufferedInputStream(new FileInputStream(new File(s)), BS);
            } catch (Exception e) {
                in = new BufferedInputStream(System.in, BS);
            }
        }

        private char getChar() {
            while (bId == size) {
                try {
                    size = in.read(buf);
                } catch (Exception e) {
                    return NC;
                }
                if (size == -1) return NC;
                bId = 0;
            }
            return (char) buf[bId++];
        }

        public int nextInt() {
            return (int) nextLong();
        }

        public int[] nextInts(int N) {
            int[] res = new int[N];
            for (int i = 0; i < N; i++) {
                res[i] = (int) nextLong();
            }
            return res;
        }

        public long[] nextLongs(int N) {
            long[] res = new long[N];
            for (int i = 0; i < N; i++) {
                res[i] = nextLong();
            }
            return res;
        }

        public long nextLong() {
            cnt = 1;
            boolean neg = false;
            if (c == NC) c = getChar();
            for (; (c < '0' || c > '9'); c = getChar()) {
                if (c == '-') neg = true;
            }
            long res = 0;
            for (; c >= '0' && c <= '9'; c = getChar()) {
                res = (res << 3) + (res << 1) + c - '0';
                cnt *= 10;
            }
            return neg ? -res : res;
        }

        public double nextDouble() {
            double cur = nextLong();
            return c != '.' ? cur : cur + nextLong() / cnt;
        }

        public double[] nextDoubles(int N) {
            double[] res = new double[N];
            for (int i = 0; i < N; i++) {
                res[i] = nextDouble();
            }
            return res;
        }

        public String next() {
            StringBuilder res = new StringBuilder();
            while (c <= 32) c = getChar();
            while (c > 32) {
                res.append(c);
                c = getChar();
            }
            return res.toString();
        }

        public String nextLine() {
            StringBuilder res = new StringBuilder();
            while (c <= 32) c = getChar();
            while (c != '\n') {
                res.append(c);
                c = getChar();
            }
            return res.toString();
        }

        public boolean hasNext() {
            if (c > 32) return true;
            while (true) {
                c = getChar();
                if (c == NC) return false;
                else if (c > 32) return true;
            }
        }
    }
    static void ASSERT(boolean assertion, String message) {
        if (!assertion) throw new AssertionError(message);
    }
    static void ASSERT(boolean assertion) {
        if (!assertion) throw new AssertionError();
    }
}