Problem: This is an interactive problem.

There is a secret permutation p (1-indexed) of numbers from 1 to n. More formally, for 1 ≤ i ≤ n, 1 ≤ p[i] ≤ n and for 1 ≤ i < j ≤ n, p[i] ≠ p[j]. It is known that p[1]<p[2].

In 1 query, you give 3 distinct integers a,b,c (1 ≤ a,b,c ≤ n), and receive the median of \{|p[a]-p[b]|,|p[b]-p[c]|,|p[a]-p[c]|\}.

In this case, the median is the 2-nd element (1-indexed) of the sequence when sorted in non-decreasing order. The median of \{4,6,2\} is 4 and the median of \{0,123,33\} is 33.

Can you find the secret permutation in not more than 2n+420 queries?

Note: the grader is not adaptive: the permutation is fixed before any queries are made.

Input

The first line of input contains a single integer t (1 ≤ t ≤ 1000) — the number of testcases.

The first line of each testcase consists of a single integer n (20 ≤ n ≤ 100000) — the length of the secret permutation.

It is guaranteed that the sum of n over all test cases does not exceed 100000.

Interaction

For each testcase, you begin the interaction by reading n.

To perform a query, output "? a b c" where a,b,c is the 3 indices you want to use for the query.

Numbers have to satisfy 1 ≤ a,b,c ≤ n and a ≠ b,b ≠ c,a ≠ c.

For each query, you will receive a single integer x: the median of \{|p[a]-p[b]|,|p[b]-p[c]|,|p[a]-p[c]|\}.

In case your query is invalid or you asked more than 2n+420 queries, the interactor will print "−1" and will finish interaction. You will receive Wrong answer verdict. Make sure to exit immediately to avoid getting other verdicts.

When you have determined the secret permutation, output "! p[1] p[2] ... p[n]". If the secret permutation is correct, the interactor will print "1". Otherwise, the interactor will print "-1" and will finish interaction. You will receive Wrong answer verdict. Make sure to exit immediately to avoid getting other verdicts.

After printing a query do not forget to output the end of line and flush the output. Otherwise, you will get Idleness limit exceeded verdict.

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 documentation for other languages.



Hacks:

To hack, use the following format of test:

The first line should contain a single integer t (1 ≤ t ≤ 1000) — the number of testcases.

The first line of each testcase should contain a single integer n (20 ≤ n ≤ 100000) — the length of the secret permutation.

The following line of should contain n integers p[1],p[2],p[3],…,p[n]. p[1]<p[2] and p must be a permutation of integers from 1 to n.

You must ensure that the sum of n over all testcases does not exceed 100000.

Example

Input


1
20

6

9

1

Output




? 1 5 2

? 20 19 2

! 9 10 19 7 16 18 11 14 15 6 20 8 17 4 5 3 12 2 13 1

Note

The secret permutation is \{9,10,19,7,16,18,11,14,15,6,20,8,17,4,5,3,12,2,13,1\}.

For the first query, the values of (a,b,c) is (1,5,2). Since p[1]=9, p[5]=16 and p[2]=10. The return value is the median of \{|9-16|,|16-10|,|9-10|\} which is 6.

For the second query, the values of (a,b,c) is (20,19,2). Since p[20]=1, p[19]=13 and p[2]=10. The return value is the median of \{|1-13|,|13-10|,|1-10|\} which is 9.

By some miracle, we have figured out that the secret permutation is \{9,10,19,7,16,18,11,14,15,6,20,8,17,4,5,3,12,2,13,1\}. We output it and receive 1 from the interactor, meaning that we have guessed the secret permutation correctly.

Solution (in CPP):

Solution:
#pragma GCC optimize ("O3")
#pragma GCC optimize ("unroll-loops")

#include<bits/stdc++.h>
#define X first
#define Y second
#define sz(a) (int)a.size()
#define ll long long
#define int long long
using namespace std;
const int N = 2e6 + 1;
int query(int a, int b, int c)
{
    cout << "? " << a << " " << b << " " << c << endl;
    int x;
    cin >> x;
    return x;
}

signed main()
{
   // ifstream cin("rt.txt.txt");
 //   ios_base::sync_with_stdio(0);
   // cin.tie(0);
   // cout.tie(0);
    int t;
    cin >> t;
    while(t--)
    {
        int n;
        cin >> n;
        vector <int> ans(n + 1);
        int a = -1, b, c;
        for(int x = 1; x <= 13; x++)
        {
            for(int y = x + 1; y <= 13; y++)
            {
                for(int z = y + 1; z <= 13; z++)
                {
                    if(query(x, y, z) <= (n - 4) / 6)
                    {
                        a = x;
                        b = y;
                        c = z;
                        break;
                    }
                }
                if(a != -1)
                {
                    break;
                }
            }
            if(a != -1)
            {
                break;
            }
        }
        map <int, vector <int> > mp;
        for(int x = 1; x <= n; x++)
        {
            if(x != a && x != b)
            {
                mp[query(a, b, x)].push_back(x);
            }
        }
        int hv;
        int largest = 0;
        for(auto it : mp)
        {
            hv = it.X;
            largest = it.Y[0];
        }
        if(sz(mp[hv - 1]) >= 2)
        {
            if(query(mp[hv][0], mp[hv - 1][0], a) > query(mp[hv][0], mp[hv -1][1], a))
            {
                swap(mp[hv - 1][0], mp[hv - 1][1]);
            }
        }
        int h1 = mp[hv][0], h2 = mp[hv - 1][0];
        ans[h1] = 1;
        ans[h2] = 2;
        for(int i = 1; i <= n; i++)
        {
            if(i != h1 && i != h2)
            {
                ans[i] = query(h1, h2, i) + 2;
            }
        }
        if(ans[1] > ans[2])
        {
            for(int i = 1; i <= n; i++)
            {
                ans[i] = n - ans[i] + 1;
            }
        }
        cout << "! ";
        for(int i = 1; i <= n; i++)
        {
            cout << ans[i] << " ";
        }
        cout << endl;
        int k;
        cin >> k;
    }
    return 0;
}