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 Solution {
    static class FastReader {   
        BufferedReader br;
        StringTokenizer st;

        public FastReader(String s) {
            try {
                br = new BufferedReader(new FileReader(s));
            } catch (FileNotFoundException e) {
                // TODO Auto-generated catch block
                e.printStackTrace();
            }
        }

        public FastReader() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        String nextToken() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    // TODO Auto-generated catch block
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(nextToken());
        }

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

        long nextLong() {
            return Long.parseLong(nextToken());
        }

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

        float nextFloat() {
            return Float.parseFloat(nextToken());
        }
    }


    static FastReader f = new FastReader();
    static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    static StringTokenizer st;
    static StringBuilder sb = new StringBuilder();
    static long[] fact;

    static int[] inputArray(int n) throws IOException {
        int[] a = new int[n];
        for(int i = 0 ; i < n ; i++) {
//            a[i] = (int) (Math.random()*1e5 + 1);
            a[i] = f.nextInt();
        }
        return a;
    }

    static void print2DArray(int[][] a) {
        int n = a.length, m = a[0].length;
        for(int i = 0 ; i < n ; i++) {
            for(int j = 0 ; j < m ; j++) {
                System.out.print(a[i][j] + " ");
            }
            System.out.println();
        }

    }

    static void print2DArray(long[][] a) {
        int n = a.length, m = a[0].length;
        for(int i = 0 ; i < n ; i++) {
            for(int j = 0 ; j < m ; j++) {
                System.out.print(a[i][j] + " ");
            }
            System.out.println();
        }

    }

    static long[] inputLongArray(int n) throws IOException {
        long[] a = new long[n];
        for(int i = 0 ; i < n ; i++) {
            a[i] = f.nextLong();
//            a[i] = (long) (Math.random() * 1e9 + 1);
        }
        return a;
    }

    static long gcd(long a , long b) {
        if(a == 0 || b == 0) {
            return Math.max(a , b);
        }
        //System.out.println("a - " + a + " b - " + b);
        if(a % b == 0) {
            return b;
        }
        return gcd(b , a % b);
    }

    static void initializeFact() {
        fact = new long[MAX_N];
        for(int i = 0 ; i < fact.length ; i++) {
            if(i == 0) {
                fact[i] = 1;
            }
            else {
                fact[i] = fact[i-1] * i % mod;
            }
        }
    }

    static long longModulus(long x , long m) {
        if(x < m) {
            return x;
        }
        long d = x / m;
        return x - d * m;
    }


    static BitSet sieveOfEratosthenes(int n)
    {
        BitSet isPrime = new BitSet(n+1);
        isPrime.set(0, n + 1);
        isPrime.set(0);
        isPrime.set(1);
        for(int i = 2; i * i <= n ; i++)
        {
            if(isPrime.get(i))
                for(int j = i * i ; j <= n; j += i)
                    isPrime.clear(j);
        }
        return isPrime;
    }

    static long moduloInversePrime(long a) {
        //System.out.println("modulo inverse of " + a + " -> " + ans);
        return modPow(a , mod - 2);
    }

    static long mult(long a, long b)
    {
        return (a * b % mod);
    }

    static long modPow(long a, int step)
    {
        long ans = 1;
        while(step != 0)
        {
            if((step & 1) != 0)
                ans = mult(ans , a);
            a = mult(a , a);
            step >>= 1;
        }
        return ans;
    }

    static int query(int t , int i, int j, int x) {
        System.out.println("? " + t + " " + i + " " + j + " " + x);
        System.out.flush();
        return f.nextInt();
    }


    private static int mod = (int) (1e9 + 7);
    static int MAX_N = (int) Math.sqrt(1e9);

    public static void main(String[] args) throws IOException {
        int t = f.nextInt();
        while(t-- != 0) {
            int n = f.nextInt();

            int max_ind = getMax(n);
            int a[] = new int[n];
            a[max_ind - 1] = n;
            for(int i = 1 ; i <= n ; i++) {
                if(i == max_ind) {
                    continue;
                }
                int q = query(2, i, max_ind, 1);
                a[i-1] = q;
            }

            System.out.print("! ");

            for(int i = 0 ; i < n ; i++) {
                System.out.print(a[i] + " ");
            }

        }

    }

    private static int getMax(int n) {
        for(int i = 0 ; i < n/2 ; i++) {
            int q = query(1, 2*i + 1, 2*i + 2, n-1);
            if(q == n) {
                return 2*i + 2;
            }
            if(q == n-1) {
                int q2 = query(2,2*i + 1, 2*i + 2, n-1);
                if(q2 == n) {
                    return 2*i + 1;
                }
            }

        }

        return n;
    }
}


/*
[3,1,4,2]

t=1 : max(min(x,pi),min(x+1,pj));
t=2: min(max(x,pi),max(x+1,pj)).

 */