So the Beautiful Regional Contest (BeRC) has come to an end! n students took part in the contest. The final standings are already known: the participant in the i-th place solved p_i problems. Since the participants are primarily sorted by the number of solved problems, then p_1 ≥ p_2 ≥ ... ≥ p_n.

Help the jury distribute the gold, silver and bronze medals. Let their numbers be g, s and b, respectively. Here is a list of requirements from the rules, which all must be satisfied:

  * for each of the three types of medals, at least one medal must be awarded (that is, g>0, s>0 and b>0); 
  * the number of gold medals must be strictly less than the number of silver and the number of bronze (that is, g<s and g<b, but there are no requirements between s and b); 
  * each gold medalist must solve strictly more problems than any awarded with a silver medal; 
  * each silver medalist must solve strictly more problems than any awarded a bronze medal; 
  * each bronze medalist must solve strictly more problems than any participant not awarded a medal; 
  * the total number of medalists g+s+b should not exceed half of all participants (for example, if n=21, then you can award a maximum of 10 participants, and if n=26, then you can award a maximum of 13 participants). 



The jury wants to reward with medals the total maximal number participants (i.e. to maximize g+s+b) so that all of the items listed above are fulfilled. Help the jury find such a way to award medals.

Input

The first line of the input contains an integer t (1 ≤ t ≤ 10000) — the number of test cases in the input. Then t test cases follow.

The first line of a test case contains an integer n (1 ≤ n ≤ 4⋅10^5) — the number of BeRC participants. The second line of a test case contains integers p_1, p_2, ..., p_n (0 ≤ p_i ≤ 10^6), where p_i is equal to the number of problems solved by the i-th participant from the final standings. The values p_i are sorted in non-increasing order, i.e. p_1 ≥ p_2 ≥ ... ≥ p_n.

The sum of n over all test cases in the input does not exceed 4⋅10^5.

Output

Print t lines, the j-th line should contain the answer to the j-th test case.

The answer consists of three non-negative integers g, s, b.

  * Print g=s=b=0 if there is no way to reward participants with medals so that all requirements from the statement are satisfied at the same time. 
  * Otherwise, print three positive numbers g, s, b — the possible number of gold, silver and bronze medals, respectively. The sum of g+s+b should be the maximum possible. If there are several answers, print any of them. 

Example

Input


5
12
5 4 4 3 2 2 1 1 1 1 1 1
4
4 3 2 1
1
1000000
20
20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1
32
64 64 63 58 58 58 58 58 37 37 37 37 34 34 28 28 28 28 28 28 24 24 19 17 17 17 17 16 16 16 16 11


Output


1 2 3
0 0 0
0 0 0
2 5 3
2 6 6

Note

In the first test case, it is possible to reward 1 gold, 2 silver and 3 bronze medals. In this case, the participant solved 5 tasks will be rewarded with the gold medal, participants solved 4 tasks will be rewarded with silver medals, participants solved 2 or 3 tasks will be rewarded with bronze medals. Participants solved exactly 1 task won't be rewarded. It's easy to see, that in this case, all conditions are satisfied and it is possible to reward participants in this way. It is impossible to give more than 6 medals because the number of medals should not exceed half of the number of participants. The answer 1, 3, 2 is also correct in this test case.

In the second and third test cases, it is impossible to reward medals, because at least one medal of each type should be given, but the number of medals should not exceed half of the number of participants.
import java.io.*;
import java.util.*;

public class tr1 {
	static PrintWriter out;
	static StringBuilder sb;
	static int inf = (int) 1e9;
	static long mod = (long) 1e9 + 7;
	static int[] si;
	static ArrayList<Integer> primes;
	static HashSet<Integer> pr;
	static int n, k, m;
	static int[] in;
	static HashMap<Integer, Integer> factors;
	static HashSet<Integer> f;
	static int[] fac, a, b;
	static int[] l, r;
	static int[][] memo;
	static int[] numc;
	static ArrayList<Integer>[] ad;
	static HashMap<Integer, Integer> hm;
	static pair[] ed;
	static TreeSet<Integer>[] np;
	static TreeMap<pair, Integer> tm;
	static int[] ans, h1, h2;
	static char[] h;
	static pair[] mm;
	static boolean[] vis;

	public static void main(String[] args) throws Exception {
		Scanner sc = new Scanner(System.in);
		out = new PrintWriter(System.out); 
	    int t=sc.nextInt();
	    while(t-->0) {
		int n=sc.nextInt();
	    int []a=new int [n];
	    for(int i=0;i<n;i++)
	    	a[i]=sc.nextInt();
	    int y=a[n/2];
	    for(int i=0;i<n;i++)
	        if(y>=a[i]) {
	        	a[i]=-1;
	        }
	    int g=1;
	    int s=0;
	    int b=0;
	    int go=a[0];
	    boolean f=true;
	    int i=1;
	             while(i<n && a[i]==go) {
	            	 if(a[i]==-1)
	            		 f=false;
	            	 i++;
	            	 g++;
	            	 
	             }
	     //        System.out.println(f+" "+g);
	             int l=0;
	     while(i<n-1 && s<=g) {
	    	 if(a[i]==-1)
        		 f=false;
	    	 l=a[i];
	    	s++;
	    	i++;
	       }
	     while(i<n && a[i]==l) {
	    	 s++;
	    	 i++;
	     }
	   //  System.out.println(f+" "+s);
	     
	     l=0;
	      while(i<n ) {
	    	  if(a[i]==-1) {
         		 break;
	    	  }
	    	  b++;
	    	  l=a[i];
	    	i++;
	       }
	      while(i<n && a[i]==l) {
		    	 b++;
		    	 i++;
		     }
	      
	  //    System.out.println(f);
	    if(!f||(g>=s) || (g>=b) || g+b+s>n/2 || g==0 || b==0 || s==0) {
	    	out.println(0+" "+0+" "+0);
	    }
	    else {
	    	out.println(g+" "+s+" "+b);
	    }
	    }
		out.flush();
	}
	
	static class pair implements Comparable<pair> {
		int x;
		int y;
        int z;
 		pair(int f, int t,int u) {
			x = f;
			y = t;
			z=u;
			// num = n;
		}

		public String toString() {
			return x + " " + y+" "+z;
		}

		@Override
		public int compareTo(pair o) {
			if(x==o.x)
				if(y==o.y)
					return z-o.z;
				else
					return o.y-y;
			return x-o.x;
		}
	}

	
	static public class FenwickTree { // one-based DS

		int n;
		int[] ft;

		FenwickTree(int size) {
			n = size;
			ft = new int[n + 1];
		}

		int rsq(int b) // O(log n)
		{
			int sum = 0;
			while (b > 0) {
				sum += ft[b];
				b -= b & -b;
			} // min?
			return sum;
		}

		int rsq(int a, int b) {
			return rsq(b) - rsq(a - 1);
		}

		void point_update(int k, int val) // O(log n), update = increment
		{
			while (k <= n) {
				ft[k] += val;
				k += k & -k;
			} // min?
		}

		int point_query(int idx) // c * O(log n), c < 1
		{
			int sum = ft[idx];
			if (idx > 0) {
				int z = idx ^ (idx & -idx);
				--idx;
				while (idx != z) {
					sum -= ft[idx];
					idx ^= idx & -idx;
				}
			}
			return sum;
		}

		void scale(int c) {
			for (int i = 1; i <= n; ++i)
				ft[i] *= c;
		}

		int findIndex(int cumFreq) {
			int msk = n;
			while ((msk & (msk - 1)) != 0)
				msk ^= msk & -msk; // msk will contain the MSB of n

			int idx = 0;
			while (msk != 0) {
				int tIdx = idx + msk;
				if (tIdx <= n && cumFreq >= ft[tIdx]) {
					idx = tIdx;
					cumFreq -= ft[tIdx];
				}
				msk >>= 1;
			}
			if (cumFreq != 0)
				return -1;
			return idx;
		}
	}

	static public class SegmentTree { // 1-based DS, OOP

		int N; // the number of elements in the array as a power of 2 (i.e. after padding)
		long[] array, sTree, lazy;

		SegmentTree(long[] in) {
			array = in;
			N = in.length - 1;
			sTree = new long[N << 1]; // no. of nodes = 2*N - 1, we add one to cross out index zero
			lazy = new long[N << 1];
			build(1, 1, N);
		}

		void build(int node, int b, int e) // O(n)
		{
			if (b == e)
				sTree[node] = array[b];
			else {
				int mid = b + e >> 1;
				build(node << 1, b, mid);
				build(node << 1 | 1, mid + 1, e);
				sTree[node] = sTree[node << 1] + sTree[node << 1 | 1];
			}
		}

		void update_point(int index, int val) // O(log n)
		{
			index += N - 1;
			sTree[index] += val;
			while (index > 1) {
				index >>= 1;
				sTree[index] = sTree[index << 1] + sTree[index << 1 | 1];
			}
		}

		void update_range(int i, int j, long val) // O(log n)
		{
			update_range(1, 1, N, i, j, val);
		}

		void update_range(int node, int b, int e, int i, int j, long val) {
			if (i > e || j < b)
				return;
			if (b >= i && e <= j) {
				sTree[node] += (e - b + 1) * val;
				lazy[node] += val;
			} else {
				int mid = b + e >> 1;
				propagate(node, b, mid, e);
				update_range(node << 1, b, mid, i, j, val);
				update_range(node << 1 | 1, mid + 1, e, i, j, val);
				sTree[node] = sTree[node << 1] + sTree[node << 1 | 1];
			}

		}

		void propagate(int node, int b, int mid, int e) {
			lazy[node << 1] += lazy[node];
			lazy[node << 1 | 1] += lazy[node];
			sTree[node << 1] += (mid - b + 1) * lazy[node];
			sTree[node << 1 | 1] += (e - mid) * lazy[node];
			lazy[node] = 0;
		}

		long query(int i, int j) {
			return query(1, 1, N, i, j);
		}

		long query(int node, int b, int e, int i, int j) // O(log n)
		{
			if (i > e || j < b)
				return 0;
			if (b >= i && e <= j)
				return sTree[node];
			int mid = b + e >> 1;
			propagate(node, b, mid, e);
			long q1 = query(node << 1, b, mid, i, j);
			long q2 = query(node << 1 | 1, mid + 1, e, i, j);
			return q1 + q2;

		}

	}

	static long gcd(long a, long b) {

		if (b == 0) {
			return a;
		}
		return gcd(b, a % b);
	}


	public int[] merge(int[] d, int st, int e) {
		if (st > e)
			return null;
		if (e == st) {
			int[] ans = { d[e] };
			return ans;
		}
		int mid = (st + e) / 2;
		int[] ans = new int[e - st + 1];
		int[] ll = merge(d, st, mid);
		int[] rr = merge(d, mid + 1, e);
		if (ll == null)
			return rr;
		if (rr == null)
			return ll;
		int iver = 0;
		int idxl = 0;
		int idxr = 0;
		for (int i = st; i <= e; i++) {
			if (ll[idxl] < rr[idxr]) {

			}
		}
		return ans;
	}

	public static class pair1 implements Comparable<pair1> {
		int a;
		int idx;

		pair1(int a, int i) {
			this.a = a;
			idx = i;
		}

		public String toString() {
			return a + " " + idx;
		}

		@Override
		public int compareTo(pair1 o) {
			return o.a - a;
		}
	}

	public static class pair2 implements Comparable<pair2> {
		int a;
		int idx;

		pair2(int a, int i) {
			this.a = a;
			idx = i;
		}

		public String toString() {
			return a + " " + idx;
		}

		@Override
		public int compareTo(pair2 o) {
			// TODO Auto-generated method stub
			return idx - o.idx;
		}
	}

	static long inver(long x) {
		int a = (int) x;
		long e = (mod - 2);
		long res = 1;
		while (e > 0) {
			if ((e & 1) == 1) {
				// System.out.println(res*a);
				res = (int) ((1l * res * a) % mod);
			}
			a = (int) ((1l * a * a) % mod);
			e >>= 1;
		}
		// out.println(res+" "+x);
		return res % mod;
	}

	static int atMostSum(int arr[], int n, int k) {
		int sum = 0;
		int cnt = 0, maxcnt = 0;

		for (int i = 0; i < n; i++) {

			// If adding current element doesn't
			// cross limit add it to current window
			if ((sum + arr[i]) <= k) {
				sum += arr[i];
				cnt++;
			}

			// Else, remove first element of current
			// window and add the current element
			else if (sum != 0) {
				sum = sum - arr[i - cnt] + arr[i];
			}

			// keep track of max length.
			maxcnt = Math.max(cnt, maxcnt);
		}
		return maxcnt;
	}

	public static int[] longestSubarray(int[] inp) {
		// array containing prefix sums up to a certain index i
		int[] p = new int[inp.length];
		p[0] = inp[0];
		for (int i = 1; i < inp.length; i++) {
			p[i] = p[i - 1] + inp[i];
		}

		// array Q from the description below
		int[] q = new int[inp.length];
		q[inp.length - 1] = p[inp.length - 1];
		for (int i = inp.length - 2; i >= 0; i--) {
			q[i] = Math.max(q[i + 1], p[i]);
		}
		int a = 0;
		int b = 0;
		int maxLen = 0;
		int curr;
		int[] res = new int[] { -1, -1 };
		while (b < inp.length) {
			curr = a > 0 ? q[b] - p[a - 1] : q[b];
			if (curr >= 0) {
				if (b - a > maxLen) {
					maxLen = b - a;
					res = new int[] { a, b };
				}
				b++;
			} else {
				a++;
			}
		}
		return res;
	}

	static void factor(int n) {
		if (si[n] == n) {
			f.add(n);
			return;
		}
		f.add(si[n]);
		factor(n / si[n]);
	}

	static void seive() {
		si = new int[100001];
		primes = new ArrayList<>();
		int N = 100001;
		si[1] = 1;
		for (int i = 2; i < N; i++) {
			if (si[i] == 0) {
				si[i] = i;
				primes.add(i);
			}
			for (int j = 0; j < primes.size() && primes.get(j) <= si[i] && (i * primes.get(j)) < N; j++)
				si[primes.get(j) * i] = primes.get(j);

		}
	}

	static class unionfind {
		int[] p;
		int[] size;
           int jum;
		unionfind(int n) {
			p = new int[n];
			size = new int[n];

			for (int i = 0; i < n; i++) {
				p[i] = i;
			}
			jum=n;
			Arrays.fill(size, 1);
		}

		int findSet(int v) {
			if (v == p[v])
				return v;
			return p[v] = findSet(p[v]);
		}

		boolean sameSet(int a, int b) {
			a = findSet(a);
			b = findSet(b);
			if (a == b)
				return true;
			return false;
		}

		int max() {
			int max = 0;
			for (int i = 0; i < size.length; i++)
				if (size[i] > max)
					max = size[i];
			return max;
		}

		void combine(int a, int b) {
			a = findSet(a);
			b = findSet(b);
			if (a == b)
				return;
			jum--;
			if (size[a] > size[b]) {
				p[b] = a;
				size[a] += size[b];

			} else {
				p[a] = b;
				size[b] += size[a];
			}
		}
	}

	static class Scanner {
		StringTokenizer st;
		BufferedReader br;

		public Scanner(InputStream system) {
			br = new BufferedReader(new InputStreamReader(system));
		}

		public Scanner(String file) throws Exception {
			br = new BufferedReader(new FileReader(file));
		}

		public String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}

		public String nextLine() throws IOException {
			return br.readLine();
		}

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

		public double nextDouble() throws IOException {
			return Double.parseDouble(next());
		}

		public char nextChar() throws IOException {
			return next().charAt(0);
		}

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

		public boolean ready() throws IOException {
			return br.ready();
		}

		public void waitForInput() throws InterruptedException {
			Thread.sleep(3000);
		}
	}
}