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 final class Solution {

    static PrintWriter out = new PrintWriter(System.out);
    static FastReader in = new FastReader();
    static long mod = (long) 1e9 + 7;
    static Pair[] moves = new Pair[]{new Pair(-1, 0), new Pair(1, 0), new Pair(0, -1), new Pair(0, 1)};

    static int[] arr;

    public static void main(String[] args) {
        int t = i();
        out:
        while (t-- > 0) {
            int n = i();

            arr = new int[n];
            int idx = -1;

            for (int i = 0; i + 1 < n; i += 2) {
                int q = query2(i, i + 1, n - 1);
                if (q == n) {
                    idx = i + 1;
                    break;
                }
                if (q == n - 1) {
                    q = query(i, i + 1, n - 1);
                    if (q == n) {
                        idx = i;
                        break;
                    }
                }
            }

            if (idx == -1) {
                idx = n - 1;
            }

            arr[idx] = n;

            for (int i = 0; i < n; i++) {
                if (i == idx) {
                    continue;
                }
                arr[i] = query(i, idx, 1);
            }

            out.print("! ");
            print(arr);
            out.flush();
        }
    }

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

    static int query2(int i, int j, int x) {
        out.println("? 1 " + (i + 1) + " " + (j + 1) + " " + x);
        out.flush();
        return i();
    }

    static boolean isPS(double number) {
        double sqrt = Math.sqrt(number);
        return ((sqrt - Math.floor(sqrt)) == 0);
    }

    static int sd(long i) {
        int d = 0;
        while (i > 0) {
            d += i % 10;
            i = i / 10;
        }
        return d;
    }

    static int[] leastPrime;

    static void sieveLinear(int N) {
        int[] primes = new int[N];
        int idx = 0;
        leastPrime = new int[N + 1];
        for (int i = 2; i <= N; i++) {
            if (leastPrime[i] == 0) {
                primes[idx++] = i;
                leastPrime[i] = i;
            }
            int curLP = leastPrime[i];
            for (int j = 0; j < idx; j++) {
                int p = primes[j];
                if (p > curLP || (long) p * i > N) {
                    break;
                } else {
                    leastPrime[p * i] = p;
                }
            }
        }
    }


    static int lower(long A[], long x) {
        int l = -1, r = A.length;
        while (r - l > 1) {
            int m = (l + r) / 2;
            if (A[m] >= x) {
                r = m;
            } else {
                l = m;
            }
        }
        return r;
    }

    static int upper(long A[], long x) {
        int l = -1, r = A.length;
        while (r - l > 1) {
            int m = (l + r) / 2;
            if (A[m] <= x) {
                l = m;
            } else {
                r = m;
            }
        }
        return l;
    }

    static void swap(int A[], int a, int b) {
        int t = A[a];
        A[a] = A[b];
        A[b] = t;
    }

    static int lowerBound(int A[], int low, int high, int x) {
        if (low > high) {
            if (x >= A[high]) {
                return A[high];
            }
        }

        int mid = (low + high) / 2;

        if (A[mid] == x) {
            return A[mid];
        }

        if (mid > 0 && A[mid - 1] <= x && x < A[mid]) {
            return A[mid - 1];
        }

        if (x < A[mid]) {
            return lowerBound(A, low, mid - 1, x);
        }

        return lowerBound(A, mid + 1, high, x);
    }


    static long pow(long a, long b) {
        long pow = 1;
        long x = a;
        while (b != 0) {
            if ((b & 1) != 0) {
                pow = (pow * x) % mod;
            }
            x = (x * x) % mod;
            b /= 2;
        }
        return pow;
    }

    static boolean isPrime(long N) {
        if (N <= 1) {
            return false;
        }
        if (N <= 3) {
            return true;
        }
        if (N % 2 == 0 || N % 3 == 0) {
            return false;
        }
        for (int i = 5; i * i <= N; i = i + 6) {
            if (N % i == 0 || N % (i + 2) == 0) {
                return false;
            }
        }
        return true;
    }

    static void print(char A[]) {
        for (char c : A) {
            out.print(c);
        }
        out.println();
    }

    static void print(boolean A[]) {
        for (boolean c : A) {
            out.print(c + " ");
        }
        out.println();
    }

    static void print(int A[]) {
        for (int c : A) {
            out.print(c + " ");
        }
        out.println();
    }

    static void print(long A[]) {
        for (long i : A) {
            out.print(i + " ");
        }
        out.println();

    }

    static void print(List<Integer> A) {
        for (int a : A) {
            out.print(a + " ");
        }
    }

    static void printYes() {
        out.println("YES");
    }

    static void printNo() {
        out.println("NO");
    }

    static int i() {
        return in.nextInt();
    }

    static long l() {
        return in.nextLong();
    }

    static String s() {
        return in.nextLine();
    }

    static int[] input(int N) {
        int A[] = new int[N];
        for (int i = 0; i < N; i++) {
            A[i] = in.nextInt();
        }
        return A;
    }

    static long[] inputLong(int N) {
        long A[] = new long[N];
        for (int i = 0; i < A.length; i++) {
            A[i] = in.nextLong();
        }
        return A;
    }

    static int GCD(int a, int b) {
        if (b == 0) {
            return a;
        } else {
            return GCD(b, a % b);
        }
    }

    static long GCD(long a, long b) {
        if (b == 0) {
            return a;
        } else {
            return GCD(b, a % b);
        }
    }

}

class BIT {

    int n;
    int[] tree;

    public BIT(int n) {
        this.n = n;
        this.tree = new int[n + 1];
    }

    public static int lowbit(int x) {
        return x & (-x);
    }

    public void update(int x) {
        while (x <= n) {
            ++tree[x];
            x += lowbit(x);
        }
    }

    public int query(int x) {
        int ans = 0;
        while (x > 0) {
            ans += tree[x];
            x -= lowbit(x);
        }
        return ans;
    }

    public int query(int x, int y) {
        return query(y) - query(x - 1);
    }
}

class SegmentTree {

    long[] t;

    public SegmentTree(int n) {
        t = new long[n + n];
        Arrays.fill(t, Long.MIN_VALUE);
    }

    public long get(int i) {
        return t[i + t.length / 2];
    }

    public void add(int i, long value) {
        i += t.length / 2;
        t[i] = value;
        for (; i > 1; i >>= 1) {
            t[i >> 1] = Math.max(t[i], t[i ^ 1]);
        }
    }

    // max[a, b]
    public long max(int a, int b) {
        long res = Long.MIN_VALUE;
        for (a += t.length / 2, b += t.length / 2; a <= b; a = (a + 1) >> 1, b = (b - 1) >> 1) {
            if ((a & 1) != 0) {
                res = Math.max(res, t[a]);
            }
            if ((b & 1) == 0) {
                res = Math.max(res, t[b]);
            }
        }
        return res;
    }
}

class Pair {

    int i, j;

    Pair(int i, int j) {
        this.i = i;
        this.j = j;
    }
}

class FastReader {

    BufferedReader br;
    StringTokenizer st;

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

    String next() {
        while (st == null || !st.hasMoreElements()) {
            try {
                st = new StringTokenizer(br.readLine());
            } catch (IOException e) {
                e.printStackTrace();
            }
        }
        return st.nextToken();
    }

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

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

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

    String nextLine() {
        String str = "";
        try {
            str = br.readLine();
        } catch (IOException e) {
            e.printStackTrace();
        }
        return str;
    }


}