Problem: This is an interactive problem!

Nastia has a hidden permutation p of length n consisting of integers from 1 to n. You, for some reason, want to figure out the permutation. To do that, you can give her an integer t (1 ≤ t ≤ 2), two different indices i and j (1 ≤ i, j ≤ n, i ≠ j), and an integer x (1 ≤ x ≤ n - 1). 

Depending on t, she will answer: 

  * t = 1: max{(min{(x, p_i)}, min{(x + 1, p_j)})}; 
  * t = 2: min{(max{(x, p_i)}, max{(x + 1, p_j)})}. 



You can ask Nastia at most ⌊ \frac {3 ⋅ n} { 2} ⌋ + 30 times. It is guaranteed that she will not change her permutation depending on your queries. Can you guess the permutation?

Input

The input consists of several test cases. In the beginning, you receive the integer T (1 ≤ T ≤ 10 000) — the number of test cases.

At the beginning of each test case, you receive an integer n (3 ≤ n ≤ 10^4) — the length of the permutation p.

It's guaranteed that the permutation is fixed beforehand and that the sum of n in one test doesn't exceed 2 ⋅ 10^4.

Interaction

To ask a question, print "? t i j x" (t = 1 or t = 2, 1 ≤ i, j ≤ n, i ≠ j, 1 ≤ x ≤ n - 1) Then, you should read the answer.

If we answer with −1 instead of a valid answer, that means you exceeded the number of queries or made an invalid query. Exit immediately after receiving −1 and you will see the Wrong Answer verdict. Otherwise, you can get an arbitrary verdict because your solution will continue to read from a closed stream.

To print the answer, print "! p_1 p_2 … p_{n} (without quotes). Note that answering doesn't count as one of the ⌊ \frac {3 ⋅ n} {2} ⌋ + 30 queries.

After printing a query or printing the answer, do not forget to output end of line and flush the output. Otherwise, you will get Idleness limit exceeded. To do this, use:

  * fflush(stdout) or cout.flush() in C++; 
  * System.out.flush() in Java; 
  * flush(output) in Pascal; 
  * stdout.flush() in Python; 
  * See the documentation for other languages. 



Hacks

To hack the solution, use the following test format.

The first line should contain a single integer T (1 ≤ T ≤ 10 000) — the number of test cases.

For each test case in the first line print a single integer n (3 ≤ n ≤ 10^4) — the length of the hidden permutation p.

In the second line print n space-separated integers p_1, p_2, …, p_n (1 ≤ p_i ≤ n), where p is permutation.

Note that the sum of n over all test cases should not exceed 2 ⋅ 10^4.

Example

Input


2
4

3

2

5

3

Output


? 2 4 1 3

? 1 2 4 2

! 3 1 4 2

? 2 3 4 2

! 2 5 3 4 1

Note

Consider the first test case.

The hidden permutation is [3, 1, 4, 2].

We print: "? 2 4 1 3" and get back min{(max{(3, p_4}), max{(4, p_1)})} = 3.

We print: "? 1 2 4 2" and get back max{(min{(2, p_2)}, min{(3, p_4)})} = 2.

Consider the second test case.

The hidden permutation is [2, 5, 3, 4, 1].

We print: "? 2 3 4 2" and get back min{(max{(2, p_3}), max{(3, p_4)})} = 3.

Solution (in JAVA):

Solution:
import java.io.*;
import java.math.*;
import java.text.*;
import java.util.*;

public class Main {
	static PrintWriter out;
	static Reader in;
	public static void main(String[] args) throws IOException {
		input_output();
		Main solver = new Main();
		solver.solve();
		out.close();
		out.flush();
	}

	static long INF = (long)1e18;
	static int MAXN = (int)4e5 + 5;
	static int MOD = (int)1e9 + 7;
	static int q, t, n, m, k;
	static double pi = Math.PI;

	void solve() throws IOException {
		t = in.nextInt();

		while (t --> 0) {
			n = in.nextInt();
			int[] ans = new int[n+1];

			int val = 0;
			boolean done = false;
			if (checkOne(1, 2)) {
				val = 1;
				done = true;
			} else if (checkN(1, 2)) {
				val = n;
				done = true;
			}

			if (!done) {
				int min1 = min(1, 2),
					max1 = max(1, 2),
					min2 = min(1, 3),
					max2 = max(1, 3);

				if (min1 == max2 || min1 == min2) {
					val = min1;
				} else val = max1;
			}

			ans[1] = val;
			for (int i = 2; i <= n; i++) {
				if (val <= n/2) {
					int tmp = 0;
					if (val != n) out.println("? "+1+" "+1+" "+i+" "+(n-1)); // check if in the right side
					else out.println("? "+1+" "+i+" "+1+" "+(n-1));
					out.flush();
					tmp = in.nextInt();
					if (tmp > val) ans[i]= tmp;
					else {
						if (val != 1) out.println("? "+2+" "+i+" "+1+" "+1); // check if in the left side
						else out.println("? "+2+" "+1+" "+i+" "+1);
						out.flush();
						tmp = in.nextInt();
						ans[i] = tmp;
					}
				} else {
					int tmp = 0;
					if (val != 1) out.println("? "+2+" "+i+" "+1+" "+1); // check if in the left side
					else out.println("? "+2+" "+1+" "+i+" "+1);
					out.flush();
					tmp = in.nextInt();
					if (tmp < val) ans[i] = tmp;
					else {
						if (val != n) out.println("? "+1+" "+1+" "+i+" "+(n-1)); // check if in the right side
						else out.println("? "+1+" "+i+" "+1+" "+(n-1));
						out.flush();
						tmp = in.nextInt();
						ans[i] = tmp;
					}
				}
			}

			out.print("!");
			for (int i = 1; i <= n; i++) out.print(" "+ans[i]);
			out.println();
			out.flush();
		}
	}

	static int min(int i, int j) {
		out.println("? "+2+" "+j+" "+i+" "+1);
		out.flush();
		return in.nextInt();
	}

	static int max(int i, int j) {
		out.println("? "+1+" "+i+" "+j+" "+(n-1));
		out.flush();
		return in.nextInt();
	}

	static boolean checkOne(int i, int j) { // check if i is equal to 1
		out.println("? "+2+" "+i+" "+j+" "+1);
		out.flush();
		int qr = in.nextInt();
		return qr == 1;
	}

	static boolean checkN(int i, int j) { // check if i is equal to N
		out.println("? "+1+" "+j+" "+i+" "+(n-1));
		out.flush();
		int qr = in.nextInt();
		return qr == n;
	}

	static class Reader {
		private InputStream mIs;
		private byte[] buf = new byte[1024];
		private int curChar;
		private int numChars;

		public Reader() {
			this(System.in);
		}

		public Reader(InputStream is) {
			mIs = is;
		}

		public int read() {
			if (numChars == -1) {
				throw new InputMismatchException();

			}
			if (curChar >= numChars) {
				curChar = 0;
				try {
					numChars = mIs.read(buf);
				} catch (IOException e) {
					throw new InputMismatchException();
				}
				if (numChars <= 0) {
					return -1;
				}
			}
			return buf[curChar++];
		}

		public String nextLine() {
			int c = read();
			while (isSpaceChar(c)) {
				c = read();
			}
			StringBuilder res = new StringBuilder();
			do {
				res.appendCodePoint(c);
				c = read();
			} while (!isEndOfLine(c));
			return res.toString();
		}

		public String next() {
			int c = read();
			while (isSpaceChar(c)) {
				c = read();
			}
			StringBuilder res = new StringBuilder();
			do {
				res.appendCodePoint(c);
				c = read();
			} while (!isSpaceChar(c));
			return res.toString();
		}

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

		public long nextLong() {
			int c = read();
			while (isSpaceChar(c)) {
				c = read();
			}
			int sgn = 1;
			if (c == '-') {
				sgn = -1;
				c = read();
			}
			long res = 0;
			do {
				if (c < '0' || c > '9') {
					throw new InputMismatchException();
				}
				res *= 10;
				res += c - '0';
				c = read();
			} while (!isSpaceChar(c));
			return res * sgn;
		}

		public int nextInt() {
			int c = read();
			while (isSpaceChar(c)) {
				c = read();
			}
			int sgn = 1;
			if (c == '-') {
				sgn = -1;
				c = read();
			}
			int res = 0;
			do {
				if (c < '0' || c > '9') {
					throw new InputMismatchException();
				}
				res *= 10;
				res += c - '0';
				c = read();
			} while (!isSpaceChar(c));
			return res * sgn;
		}

		public boolean isSpaceChar(int c) {
			return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
		}

		public boolean isEndOfLine(int c) {
			return c == '\n' || c == '\r' || c == -1;
		}

	}
	static void input_output() throws IOException {
		File f = new File("in.txt");
		if (f.exists() && !f.isDirectory()) {
			in = new Reader(new FileInputStream("in.txt"));
		} else in = new Reader();
		f = new File("out.txt");
		if (f.exists() && !f.isDirectory()) {
			out = new PrintWriter(new File("out.txt"));
		} else out = new PrintWriter(System.out);
	}
}