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.*;

public class CodeForces extends Functions {
    public static void solve(){
        int n = sc.nextInt();
        int[] arr = new int[n+1];
        int oneIndex = n;
        // find one index
        for(int i=1;i<n;i+=2){
            out.println("? 2 "+i+" "+(i+1)+" "+1);
            int x = sc.nextInt();
            out.flush();
            if(x==1){
                oneIndex = i;
                break;
            }
            if(x==2){
                out.println("? 1 "+i+" "+(i+1)+" "+1);
                int temp = sc.nextInt();
                out.flush();
                if(temp==1){
                    oneIndex = i+1;
                    break;
                }
            }
        }

        arr[oneIndex] = 1;
        for(int i=1;i<=n;i++){
            if(i==oneIndex)continue;
            out.println("? 1 "+oneIndex+" "+i+" "+(n-1));
            arr[i] = sc.nextInt();
            out.flush();
        }
        StringBuilder sb = new StringBuilder();
        sb.append("!");
        for(int i=1;i<=n;i++)sb.append(" ").append(arr[i]);
        out.println(sb);
        out.flush();

    }
    public static void main(String[] args) {
        long start = System.currentTimeMillis();
        int testCase = 1;
        testCase= sc.nextInt();
        while (testCase-->0) solve();
        long end = System.currentTimeMillis();
//        System.out.println("time took in ms : "+(end-start));
        sc.close();
        out.close();
    }

    static Scanner sc = new Scanner(); // Input
    static PrintStream out = new PrintStream(System.out);  // output
}

class Functions {
    public static long mod = (long)1e9+7L;
    public static int INT_MAX = Integer.MAX_VALUE;
    public static int INT_MIN = Integer.MIN_VALUE;
    public static long LONG_MAX = Long.MAX_VALUE;
    public static long LONG_MIN = Long.MIN_VALUE;
    public static double DOUBLE_MAX = Double.MAX_VALUE;
    public static double DOUBLE_MIN = Double.MIN_VALUE;

    public static void sort(int[] a,boolean isAscending){
        ArrayList<Integer> temp = new ArrayList<>();
        for (int j : a) temp.add(j);
        sort(temp,isAscending);
        for(int i=0;i<a.length;i++)a[i] = temp.get(i);
    }

    public static void sort(List list,boolean isAscending){
        if(isAscending)
            Collections.sort(list);
        else Collections.sort(list,Collections.reverseOrder());
    }
    // euclidean algorithm
    public static long gcd(long a, long b) {
        // time O(max (loga ,logb))
        long x = Math.min(a,b);
        long y = Math.max(a,b);
        if (y % x == 0) return x;
        return gcd(y%x,x);
    }
    public static long lcm(long a, long b) {
        // lcm(a,b) * gcd(a,b) = a * b
        return (a / gcd(a, b)) * b;
    }

    public static long factorial(int n){
        long fact = 1L;
        for(int i=2;i<=n;i++)fact = (fact*i)%mod;
        return fact;
    }
    public static long isPrime(int n){
        for(long i=2;i*i<=n;i++)
            if(n%i==0)return i;
        return -1;
    }
    public static long power(long x,long n){
        // calculate x^n %mod
        // time : O(logn)
        long ans = 1;
        while(n>0){
            if(n%2!=0){
                ans *= x;
                n--;
            }else {
                x *= x;
                n/=2;
            }
            x %=mod;
            ans %=mod;
        }
        return ans;
    }

    public  static int ncr(int n, int r){
        // time O(n+r)
        if (r > n)
            return 0;
        long[] inv = new long[r + 1];
        inv[1] = 1;
        // Getting the modular inversion
        // for all the numbers
        // from 2 to r with respect to m
        for (int i = 2; i <= r; i++) {
            inv[i] = mod - (mod / i) * inv[(int) (mod % i)] % mod;
        }
        int ans = 1;
        // for 1/(r!) part
        for (int i = 2; i <= r; i++) {
            ans = (int) (((ans % mod) * (inv[i] % mod)) % mod);
        }
        // for (n)*(n-1)*(n-2)*...*(n-r+1) part
        for (int i = n; i >= (n - r + 1); i--) {
            ans = (int) (((ans % mod) * (i % mod)) % mod);
        }
        return ans;
    }

    public static void reverseArray(int[] a){
        int left = 0;
        int right = a.length-1;
        while (left<right){
            int temp =a[left];
            a[left] = a[right];
            a[right] = temp;
            left++;
            right--;
        }
    }


}

class Scanner {
    private BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    private StringTokenizer st = new StringTokenizer("");
    public String next(){
        while (!st.hasMoreTokens())
            try {
                st = new StringTokenizer(br.readLine());
            }catch (IOException e){
                e.printStackTrace();
            }
        return st.nextToken();

    }
    public int nextInt(){return Integer.parseInt(next());}
    public long nextLong(){return Long.parseLong(next());}
    public double nextDouble(){return Double.parseDouble(next());}
    public int[] setintArray(int n){
        int[] arr =new int[n];
        for(int i=0;i<n;i++)arr[i] = nextInt();
        return arr;
    }
    public Integer[] setIntegerArray(int n){
        Integer[] arr =new Integer[n];
        for(int i=0;i<n;i++)arr[i] = nextInt();
        return arr;
    }
    public long[] setlongArray(int n){
        long[] arr =new long[n];
        for(int i=0;i<n;i++)arr[i] = nextLong();
        return arr;
    }
    public int[][] set2DIntegerMatrix(int row,int col){
        int[][] arr = new int[row][col];
        for(int i=0;i<row;i++)
            for(int j= 0;j<col;j++)
                arr[i][j] = nextInt();
        return arr;
    }
    public void close(){
        try{
            br.close();
        }catch (IOException e){
            System.out.println(e.getMessage());
        }
    }
}