The only difference between easy and hard versions is constraints.

You are given a sequence a consisting of n positive integers.

Let's define a three blocks palindrome as the sequence, consisting of at most two distinct elements (let these elements are a and b, a can be equal b) and is as follows: [\underbrace{a, a, ..., a}_{x}, \underbrace{b, b, ..., b}_{y}, \underbrace{a, a, ..., a}_{x}]. There x, y are integers greater than or equal to 0. For example, sequences [], [2], [1, 1], [1, 2, 1], [1, 2, 2, 1] and [1, 1, 2, 1, 1] are three block palindromes but [1, 2, 3, 2, 1], [1, 2, 1, 2, 1] and [1, 2] are not.

Your task is to choose the maximum by length subsequence of a that is a three blocks palindrome.

You have to answer t independent test cases.

Recall that the sequence t is a a subsequence of the sequence s if t can be derived from s by removing zero or more elements without changing the order of the remaining elements. For example, if s=[1, 2, 1, 3, 1, 2, 1], then possible subsequences are: [1, 1, 1, 1], [3] and [1, 2, 1, 3, 1, 2, 1], but not [3, 2, 3] and [1, 1, 1, 1, 2].

Input

The first line of the input contains one integer t (1 ≤ t ≤ 2000) — the number of test cases. Then t test cases follow.

The first line of the test case contains one integer n (1 ≤ n ≤ 2000) — the length of a. The second line of the test case contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 26), where a_i is the i-th element of a. Note that the maximum value of a_i can be up to 26.

It is guaranteed that the sum of n over all test cases does not exceed 2000 (∑ n ≤ 2000).

Output

For each test case, print the answer — the maximum possible length of some subsequence of a that is a three blocks palindrome.

Example

Input


6
8
1 1 2 2 3 2 1 1
3
1 3 3
4
1 10 10 1
1
26
2
2 1
3
1 1 1


Output


7
2
4
1
1
3
import java.io.ByteArrayInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.InputMismatchException;
import java.util.List;

public class Main {
	InputStream is;
	PrintWriter out;
	String INPUT = "";

	void solve() {
		for(int T = ni(); T > 0; T--) {
			int n = ni();
			int[] a = na(n);
			int[][] f = new int[n][27];
			f[0][a[0]]++;
			for(int i = 1; i < n; i++) {
				for(int j = 0; j < f[i].length; j++) {
					f[i][j] += f[i-1][j];
				}
				f[i][a[i]]++;
			}
			int[][] revf = new int[n][27];
			revf[n-1][a[n-1]]++;
			for(int i = n-2; i >= 0; i--) {
				for(int j = 0; j < revf[i].length; j++) {
					revf[i][j] += revf[i+1][j];
				}
				revf[i][a[i]]++;
			}
			int max = 0;
			for(int c = 1; c < 27; c++) {
				for(int i = 0; i < n; i++) {
					if(f[i][c] > 0) {
						for(int j = i+1; j < n; j++) {
							if(f[i][c] == revf[j][c]) {
								for(int cc = 1; cc < 27; cc++) {
									int r = f[j-1][cc] - f[i][cc];
									max = Math.max(max, 2*f[i][c] + r);
								}
							}
						}
					}
				}
			}
			out.println(max == 0 ? 1 : max);
		}
	}

	
	void run() throws Exception {
		is = INPUT.isEmpty() ? System.in : new ByteArrayInputStream(INPUT.getBytes());
		out = new PrintWriter(System.out);

		long s = System.currentTimeMillis();
		solve();
		out.flush();
		if (!INPUT.isEmpty())
			tr(System.currentTimeMillis() - s + "ms");
	}

	public static void main(String[] args) throws Exception {
		new Main().run();
	}

	private byte[] inbuf = new byte[1024];
	public int lenbuf = 0, ptrbuf = 0;

	private int readByte() {
		if (lenbuf == -1)
			throw new InputMismatchException();
		if (ptrbuf >= lenbuf) {
			ptrbuf = 0;
			try {
				lenbuf = is.read(inbuf);
			} catch (IOException e) {
				throw new InputMismatchException();
			}
			if (lenbuf <= 0)
				return -1;
		}
		return inbuf[ptrbuf++];
	}

	private boolean isSpaceChar(int c) {
		return !(c >= 33 && c <= 126);
	}

	private int skip() {
		int b;
		while ((b = readByte()) != -1 && isSpaceChar(b))
			;
		return b;
	}

	private double nd() {
		return Double.parseDouble(ns());
	}

	private char nc() {
		return (char) skip();
	}

	private String ns() {
		int b = skip();
		StringBuilder sb = new StringBuilder();
		while (!(isSpaceChar(b))) { // when nextLine, (isSpaceChar(b) && b != ' ')
			sb.appendCodePoint(b);
			b = readByte();
		}
		return sb.toString();
	}

	private char[] ns(int n) {
		char[] buf = new char[n];
		int b = skip(), p = 0;
		while (p < n && !(isSpaceChar(b))) {
			buf[p++] = (char) b;
			b = readByte();
		}
		return n == p ? buf : Arrays.copyOf(buf, p);
	}

	private char[][] nm(int n, int m) {
		char[][] map = new char[n][];
		for (int i = 0; i < n; i++)
			map[i] = ns(m);
		return map;
	}

	private int[] na(int n) {
		int[] a = new int[n];
		for (int i = 0; i < n; i++)
			a[i] = ni();
		return a;
	}

	private int ni() {
		int num = 0, b;
		boolean minus = false;
		while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'))
			;
		if (b == '-') {
			minus = true;
			b = readByte();
		}

		while (true) {
			if (b >= '0' && b <= '9') {
				num = num * 10 + (b - '0');
			} else {
				return minus ? -num : num;
			}
			b = readByte();
		}
	}

	private long nl() {
		long num = 0;
		int b;
		boolean minus = false;
		while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'))
			;
		if (b == '-') {
			minus = true;
			b = readByte();
		}

		while (true) {
			if (b >= '0' && b <= '9') {
				num = num * 10 + (b - '0');
			} else {
				return minus ? -num : num;
			}
			b = readByte();
		}
	}

	private static void tr(Object... o) {
		System.out.println(Arrays.deepToString(o));
	}
}