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.*;
import java.util.*;


public class Solution {
  public static void solve(BufferedReader br) throws IOException {
    int n = Integer.parseInt(br.readLine());
    String[] s = br.readLine().split(" ");
    int[] arr = new int[n];
    for (int i=0; i<n; i++) {
      arr[i] = Integer.parseInt(s[i]);
    }

    // construct prefix and suffix max
    int[] prefix_max = new int[n];
    int[] suffix_max = new int[n];

    prefix_max[0] = arr[0];
    for (int i=1; i<n; i++) {
      prefix_max[i] = Math.max(prefix_max[i-1], arr[i]);
    }
    suffix_max[n-1] = arr[n-1];
    for (int i=n-2; i>=0; i--) {
      suffix_max[i] = Math.max(suffix_max[i+1], arr[i]);
    }

    // find maximum value and their indices in the array
    List<Integer> list = new ArrayList<>();
    for (int i=0; i<n; i++) {
      if (arr[i] == prefix_max[n-1]) {
        list.add(i);
      }
    }

    // let l = r be left and right bound of the middle partition, and set l = r be l.get(l.size() / 2)
    // which means set l be the middle position of maximum values, we notice that if there are at least 3
    // maximum values, it is always posible for the paritition. Elsewise, we expand our bound to the left or 
    // to the right, determined by which side's expanding making it more likely to satisfy condition.
    int l = list.get(list.size()/2);
    int r = l;
    int middle_min = prefix_max[n-1]; // starting with single element, which is also the maximum element in the array
    while (true) {
      // if we reach to the bound of the array, print NO
      if (l == 0 || r == n-1) {
        System.out.println("NO");
        return;
      }
    
      // otherwise we firstt check if we meet the condition
      if (prefix_max[l-1] == middle_min && suffix_max[r+1] == middle_min) {
        System.out.println("YES");
        System.out.format("%d %d %d%n", l, r-l+1, n-r-1);
        return;
      }
      
      // if not meet, we move l to the left or r to the right
      // we greedily pick a side that causes min(left_max, middle_min, right_max) decrease minimum
      // due to the fact that we always decrease left_max, middle_min and right_max by expanding our middle partition
      int left_max = (l - 1 == 0) ? Integer.MIN_VALUE : Math.min(prefix_max[l-2], arr[l-1]); // new min(u,v,w) is determined by prefix[l-2] and arr[l-1]
      int right_max = (r + 1 == n-1) ? Integer.MIN_VALUE : Math.min(suffix_max[r+2], arr[r+1]); // new min(u,v,w) is determined by suffixe[r+2] and arr[r+1]
      if (left_max >= right_max) {
        // it is better to expand middle partition to the left
        middle_min = Math.min(middle_min, arr[l-1]);
        l--;
      } else {
        // it is better to expand middle partition to the right
        middle_min = Math.min(middle_min, arr[r+1]);
        r++;
      }
    }

  }
  public static void main(String[] args) throws IOException {
    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    int t = Integer.parseInt(br.readLine());
    while (t-- > 0) {
      solve(br);
    }
  }
}