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.awt.Point;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.io.ObjectInputStream.GetField;
import java.lang.reflect.Array;
import java.math.BigDecimal;
import java.math.BigInteger;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collection;
import java.util.Collections;
import java.util.HashSet;
import java.util.Hashtable;
import java.util.LinkedList;
import java.util.PriorityQueue;
import java.util.Queue;
import java.util.Set;
import java.util.Stack;
import java.util.StringTokenizer;
import java.util.TreeMap;
import java.util.TreeSet;

import javax.sound.sampled.ReverbType;


public class N720 {
	static PrintWriter out;
	static Scanner sc;
	static ArrayList<int[]>q,w,x;
	static ArrayList<Integer>adj[];
	static HashSet<Integer>primesH;
	static boolean prime[];
	//static ArrayList<Integer>a;
	static HashSet<Long>tmp;
	static int[][][]dist;
	static boolean[]v;
	static int[]a,b,c,d;
	static Boolean[][]dp;
	static char[][]mp;
	static int A,B,n,m,h,ans,sum;
	//static String a,b;
	static long oo=(long)1e9+7;
	public static void main(String[]args) throws IOException {
		sc=new Scanner(System.in);
		out=new PrintWriter(System.out);
		//A();
		//B();
		C();
		//D();
		//E();
		//F();
		//G();
		out.close();
	}
	  private static void A() throws IOException {
		  int t=ni();
		  while(t-->0) {
				long a=nl(),b=nl();
				if(b==1) {
					ol("NO");continue;
				}
				ol("YES");
				if(b==2) {
					ol(a+" "+(2l*a)+" "+(3l*a));
					continue;
				}
				ol((a*1l*(b-1))+" "+(a)+" "+(a*1l*b));
		  }
		
	   }
		static void B() throws IOException {
			int t=ni();
			while(t-->0) {
				int n=ni();
				a=nai(n);
				int mn=Integer.MAX_VALUE,idx=0;
				for(int i=0;i<n;i++) {
					if(a[i]<mn) {
						mn=a[i];
						idx=i;
					}
				}
				ol(n-1);
				boolean x=true;
				for(int i=idx-1;i>=0;i--) {
					if(x) {
						ol((idx+1)+" "+(i+1)+" "+(mn)+" "+(mn+1));
					}else {
						ol((idx+1)+" "+(i+1)+" "+(mn)+" "+(mn));
					}
					x=!x;
				}
				x=true;
				for(int i=idx+1;i<n;i++) {
					if(x) {
						ol((idx+1)+" "+(i+1)+" "+(mn)+" "+(mn+1));
					}else {
						ol((idx+1)+" "+(i+1)+" "+(mn)+" "+(mn));
					}
					x=!x;
				}
			}
		}
		
		static void C() throws IOException{
			 int t=ni();
			 while(t-->0) {
				 int n=ni();
				 int[]ans=new int[n];
				 boolean[]v=new boolean[n+1];
				 int k=n;if(k%2==1)k--;
				 for(int i=0;i<k;i+=2) {
					 out.println("? 1 "+(i+1)+" "+(i+2)+" "+(n-1));
					 out.flush();
					 int mx;
					 int val=ni();
					 mx=val;
					 if(val==n-1) {
						 out.println("? 1 "+(i+2)+" "+(i+1)+" "+(n-1));
						 out.flush();
						 int tmp=ni();
						 if(tmp==n)mx=n;
					 }
					 out.println("? 1 "+(i+1)+" "+(i+2)+" "+(mx-1));
					 out.flush();
					 int tmp=ni();
					 if(tmp==mx-1) {//pi = mx
						 ans[i]=mx;
						 out.println("? 2 "+(i+2)+" "+(i+1)+" "+1);
						 out.flush();
						 tmp=ni();
						 ans[i+1]=tmp;
					 }else {
						 ans[i+1]=mx;
						 out.println("? 2 "+(i+1)+" "+(i+2)+" "+1);
						 out.flush();
						 tmp=ni();
						 ans[i]=tmp;
					 }
					 v[mx]=v[tmp]=true;
				 }
				 if(n%2==1) {
					 for(int i=1;i<=n;i++) {
						 if(!v[i]) {
							 ans[n-1]=i;break;
						 }
					 }
				 }
				 out.print("!");
				 for(int i=0;i<ans.length;i++) {
					 out.print(" "+ans[i]);
				 }
				 out.println();out.flush();
			 }
		}
		private static Boolean dp(int i, int j) {
			if(j>sum/2)return false;
			if(i==x.size()) {
				return sum/2==j;
			}
			if(dp[i][j]!=null)return dp[i][j];
			
			return dp[i][j]=dp(i+1,j+x.get(i)[0])||dp(i+1,j);
		}
		static boolean isPrime(long n) {
			if(n==2)return true;
			if(n<2||n%2==0)return false;
			
			for(long i=3L;i*i<n;i+=2l) {
				long rem=(n%i);
				if(rem==0)return false;
			}
			return true;
		}
		static void D() throws IOException {
			int t=1;
			while(t-->0) {
				int n=ni(),m=ni(),k=ni();
				int[][]ans=new int[n][m];
				dist=new int[n][m][4];
				for(int i=0;i<n;i++)for(int j=0;j>m;j++)
					Arrays.fill(dist[i][j], Integer.MAX_VALUE);
				int x;
				for(int i=0;i<n;i++) {
					for(int j=0;j<m-1;j++) {
						dist[i][j][2]=(x=ni());
						dist[i][j+1][3]=x;
					}
				}
				for(int i=0;i<n-1;i++) {
					for(int j=0;j<m;j++) {
						dist[i][j][1]=(x=ni());
						dist[i+1][j][0]=x;
					}
				}
				int[][]nans=new int[n][m];
				if(k%2==1) {
					for(int i=0;i<n;i++)Arrays.fill(ans[i], -1);
				}else {
					for(int ii=0;ii<k/2;ii++) {
					for(int i=0;i<n;i++) {
						
						for(int j=0;j<m;j++) {
							nans[i][j]=Integer.MAX_VALUE;
							if(i>0)
								nans[i][j]=Math.min(nans[i][j], ans[i-1][j]+dist[i-1][j][1]);
							if(i<n-1)
								nans[i][j]=Math.min(nans[i][j], ans[i+1][j]+dist[i+1][j][0]);
							if(j>0)
								nans[i][j]=Math.min(nans[i][j], ans[i][j-1]+dist[i][j-1][2]);
							if(j<m-1)
								nans[i][j]=Math.min(nans[i][j], ans[i][j+1]+dist[i][j+1][3]);
						}
					}
					int[][]tmp=ans;
					ans=nans;
					nans=tmp;
					}
				}
				for(int i=0;i<n;i++) {
					for(int j=0;j<m;j++) {
						if(ans[i][j]!=-1)ans[i][j]*=2;
						out.print(ans[i][j]+" ");
					}
					ol("");
				}
			}
		}
		private static int bfs(int i, int j,int k) {
			boolean [][]vis=new boolean[dist.length][dist[0].length];
			Queue<int[]>q=new LinkedList<int[]>();
			int mn=Integer.MAX_VALUE;
			q.add(new int[] {i,j,0,0});
			int[]dx=new int[] {-1,1,0,0};
			int[]dy=new int[] {0,0,1,-1};
			while(!q.isEmpty()) {
				int []x=q.poll();
				vis[x[0]][x[1]]=true;
				int c=x[2];
				if(c>k/2)continue;
				if(c>0&&k%c==0&&(k/c)%2==0) {
					mn=Math.min(mn,x[3]*k/c );
				}
				for(int a=0;a<4;a++) {
					int nx=x[0]+dx[a];
					int ny=x[1]+dy[a];
					if(valid(nx,ny)&&!vis[nx][ny]) {
						q.add(new int[] {nx,ny,c+1,x[3]+dist[x[0]][x[1]][a]});
					}
				}
				
			}
			return mn;
		}
		private static boolean valid(int nx, int ny) {
			return nx>=0&&nx<dist.length&&ny>=0&&ny<dist[0].length;
		}
		static int gcd (int a, int b) {
		    return b==0?a:gcd (b, a % b);
		}
		
	static void E() throws IOException {
		int t=ni();
		while(t-->0) {
			
			
		}
		
	}

	static void F() throws IOException {
		int t=ni();
		while(t-->0) {
			
		}
	}
	static void CC() throws IOException {
		for(int kk=2;kk<21;kk++) {
			ol(kk+"   -------");
		int n=kk;
		int k=n-2;
		int msk=1<<k;
		int[]a=new int[k];
		for(int i=0;i<a.length;i++)a[i]=i+2;
		int mx=1;
		int ms=0;
		for(int i=1;i<msk;i++) {
			long prod=1;
			int cnt=0;
			for(int j=0;j<a.length;j++) {
				if(((i>>j)&1)!=0) {
					prod*=a[j];
					cnt++;
				}
			}
			if(cnt>=mx&&prod%n==1) {
				mx=cnt;
				ms=i;
			}
			
		}
		ol(mx==1?mx:mx+1);
		out.print(1+" ");
		long pr=1;
		for(int j=0;j<a.length;j++) {
			if(((ms>>j)&1)!=0) {
				out.print(a[j]+" ");
				pr*=a[j];
			}
		}
		ol("");
		ol("Prod: "+pr);
		ol(n+"*"+((pr-1)/n)+" + "+1);
		}
	}
	static int ni() throws IOException {
		return sc.nextInt();
	}
	static double nd() throws IOException {
		return sc.nextDouble();
	}
	static long nl() throws IOException {
		return  sc.nextLong();
	}
	static String ns() throws IOException {
		return sc.next();
	}
	static int[] nai(int n) throws IOException {
		int[] a = new int[n];
		for (int i = 0; i < n; i++)
			a[i] = sc.nextInt();
		return a;
	}
	static long[] nal(int n) throws IOException {
		long[] a = new long[n];
		for (int i = 0; i < n; i++)
			a[i] = sc.nextLong();
		return a;
	}
	static int[][] nmi(int n,int m) throws IOException{
		int[][]a=new int[n][m];
		for(int i=0;i<n;i++) {
			for(int j=0;j<m;j++) {
				a[i][j]=sc.nextInt();
			}
		}
		return a;
	}

	static long[][] nml(int n,int m) throws IOException{
		long[][]a=new long[n][m];
		for(int i=0;i<n;i++) {
			for(int j=0;j<m;j++) {
				a[i][j]=sc.nextLong();
			}
		}
		return a;
	}
	static void o(String x) {
		out.print(x);
	}
	static void ol(String x) {
		out.println(x);
	}
	static void ol(int x) {
		out.println(x);
	}
	static void disp1(int []a) {
		for(int i=0;i<a.length;i++) {
			out.print(a[i]+" ");
		}
		out.println();
	}

	static class Scanner 
	{
		StringTokenizer st;
		BufferedReader br;

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

		public String next() throws IOException 
		{
			while (st == null || !st.hasMoreTokens()) 
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}
		public boolean hasNext() {return st.hasMoreTokens();}
		public int nextInt() throws IOException {return Integer.parseInt(next());}
		
		public double nextDouble() throws IOException {return Double.parseDouble(next());}
		
		public long nextLong() throws IOException {return Long.parseLong(next());}

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

	}
}