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.util.*;
 
import java.math.*;
import java.awt.Point;
 
public class Main {
    //static final long MOD = 1000000007L;
    //static final long MOD2 = 1000000009L;
    static final long MOD = 998244353L;
    //static final long INF = 500000000000L;
    static final int INF =   1000000005;
    static final int NINF = -1000000005;
    //static final long NINF = -1000000000000000000L;
    static FastScanner sc;
    static PrintWriter pw;
    static final int[][] dirs = {{-1,0},{1,0},{0,-1},{0,1}};
 
    static int N;
 
    public static void main(String[] args) {
        sc = new FastScanner();
        pw = new PrintWriter(System.out);
 
        int Q = sc.ni();
        for (int q = 0; q < Q; q++) {
            N = sc.ni();
            
            int max01 = max(0,1);
 
            int max12 = max(1,2);
 
            int min01 = min(0,1);
 
            int min12 = min(1,2);
 
            int a1;
            if (max01==max12 || max01==min12)
                a1 = max01;
            else
                a1 = min01;
            
            int[] ans = new int[N];
            ans[0] = other(min01,max01,a1);
            ans[1] = a1;
            ans[2] = other(min12,max12,a1);
 
            BinaryIndexedTree bit = new BinaryIndexedTree(N);
            for (int i = 0; i < N; i++)
                bit.add(1, i);
            bit.add(-1,ans[0]-1);
            bit.add(-1,ans[1]-1);
            bit.add(-1,ans[2]-1);
 
            for (int i = 2; i < N-1; i++) {
                int less = bit.sum(0,ans[i]-1);
                int total = bit.sum(0,N-1);
                if (less >= total/2) {
                    int min = min(i,i+1);
                    if (min < ans[i]) {
                        ans[i+1] = min;
                    } else {
                        int max = max(i,i+1);
                        ans[i+1] = max;
                    }
                } else {
                    int max = max(i,i+1);
                    if (max > ans[i]) {
                        ans[i+1] = max;
                    } else {
                        int min = min(i,i+1);
                        ans[i+1] = min;
                    }
                }
                bit.add(-1,ans[i+1]-1);
            }
 
            StringBuilder a = new StringBuilder();
            a.append("! ");
            for (int n: ans) {
                a.append(n + " ");
            }
            pw.println(a.toString());
            pw.flush();
        }
        pw.close();
    }
 
    static class BinaryIndexedTree {
        public int[] arr;
      
        public BinaryIndexedTree (int N) {
          arr = new int[N+1];
        }
      
        //add k to the i-th element.
        public void add(int k, int i) {
          int node = i+1;
          while (node < arr.length) {
            arr[node] += k;
            node += node & (-node);
          }
        }
      
        //sum up the elements from input[s_i] to input[e_i], from [s_i,e_i].
        public int sum(int s_i, int e_i) {
          return sum(e_i+1) - sum(s_i);
        }
      
        private int sum(int i) {
          int total = 0;
          int node = i;
          while (node > 0) {
            total += arr[node];
            node -= node & (-node);
          }
          return total;
        }
    }
 
    public static int max(int i, int j) {
        pw.println("? 1 " + (i+1) + " " + (j+1) + " " + (N-1)); //max(A[i],A[j])
        pw.flush();
        int max = sc.ni();
        if (max==N-1) {
            //could be N
            pw.println("? 1 " + (j+1) + " " + (i+1) + " " + (N-1));
            pw.flush();
            int max2 = sc.ni();
            if (max2==N) {
                return N;
            }
        }
        return max;
    }
 
    public static int min(int i, int j) {
        pw.println("? 2 " + (i+1) + " " + (j+1) + " " + 1); //min(A[i],A[j])
        pw.flush();
        int min = sc.ni();
        if (min==2) {
            //could be 1
            pw.println("? 2 " + (j+1) + " " + (i+1) + " " + 1);
            pw.flush();
            int min2 = sc.ni();
            if (min2==1) {
                return 1;
            }
        }
        return min;
    }
 
    public static int other(int a, int b, int notX) {
        if (a==notX)
            return b;
        else
            return a;
    }
 
    public static long gcd(long a, long b) {
        if (a < b)
          return gcd(b,a);
        if (b == 0)
          return a;
        else
          return gcd(b,a%b);
    }
 
    public static boolean isSquare(long x) {
        long s = (long)Math.round(Math.sqrt(x));
        return s*s==x;
    }
 
 
    public static void sort(int[] arr) {
        Random rgen = new Random();
        for (int i = 0; i < arr.length; i++) {
            int r = rgen.nextInt(arr.length);
            int temp = arr[i];
            arr[i] = arr[r];
            arr[r] = temp;
        }
        Arrays.sort(arr);
    }
 
    public static void sort(long[] arr) {
        Random rgen = new Random();
        for (int i = 0; i < arr.length; i++) {
            int r = rgen.nextInt(arr.length);
            long temp = arr[i];
            arr[i] = arr[r];
            arr[r] = temp;
        }
        Arrays.sort(arr);
    }
 
    //Sort an array (immune to quicksort TLE)
    public static void sort(int[][] arr) {
        Random rgen = new Random();
        for (int i = 0; i < arr.length; i++) {
            int r = rgen.nextInt(arr.length);
            int[] temp = arr[i];
            arr[i] = arr[r];
            arr[r] = temp;
        }
        Arrays.sort(arr, new Comparator<int[]>() {
            @Override
            public int compare(int[] a, int[] b) {
                return a[0]-b[0];
            }
        });
    }
    
    public static void sort(long[][] arr) {
        Random rgen = new Random();
        for (int i = 0; i < arr.length; i++) {
            int r = rgen.nextInt(arr.length);
            long[] temp = arr[i];
            arr[i] = arr[r];
            arr[r] = temp;
        }
        Arrays.sort(arr, new Comparator<long[]>() {
            @Override
            public int compare(long[] a, long[] b) {
                if (a[0] > b[0])
                    return 1;
                else if (a[0] < b[0])
                    return -1;
                else
                    return 0;
                //Ascending order.
            }
        });
    }
 
    static class FastScanner {
        BufferedReader br;
        StringTokenizer st;
 
        public FastScanner() {
            br = new BufferedReader(new InputStreamReader(System.in), 32768);
            st = null;
        }
 
        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }
 
        int ni() {
            return Integer.parseInt(next());
        }
 
        int[][] graph(int N, int[][] edges) {
            int[][] graph = new int[N][];
            int[] sz = new int[N];
            for (int[] e: edges) {
                sz[e[0]] += 1;
                sz[e[1]] += 1;
            }
            for (int i = 0; i < N; i++) {
                graph[i] = new int[sz[i]];
            }
            int[] cur = new int[N];
            for (int[] e: edges) {
                graph[e[0]][cur[e[0]]] = e[1];
                graph[e[1]][cur[e[1]]] = e[0];
                cur[e[0]] += 1;
                cur[e[1]] += 1;
            }
            return graph;
        }
 
        int[] intArray(int N, int mod) {
            int[] ret = new int[N];
            for (int i = 0; i < N; i++)
                ret[i] = ni()+mod;
            return ret;
        }
 
        long nl() {
            return Long.parseLong(next());
        }
 
        long[] longArray(int N, long mod) {
            long[] ret = new long[N];
            for (int i = 0; i < N; i++)
                ret[i] = nl()+mod;
            return ret;
        }
 
        double nd() {
            return Double.parseDouble(next());
        }
 
        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }
}