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 CPP):

Solution:
//DAILY HUSTLE MORE
#include<bits/stdc++.h>
using namespace std;
#define   int             long long               // ll
#define   f(a,b,n)        for(ll a=b;a<n;a++)
#define   readarr(arr,n)  f(i,0,n) cin>>arr[i];
#define   ll              long long
#define   pb              push_back
#define   go              ios_base::sync_with_stdio(false),cin.tie(NULL),cout.tie(NULL)
#define   ms(a,i)         memset(a,i,sizeof(a))
#define   mod             1000000007 // mod 
#define   w(t)            int t=1; cin>>t; while(t--)
#define   MOD             1e9+7
#define   setbits(x)      __builtin_popcountll(x)
#define   zrobits(x)      __builtin_ctzll(x)
#define   all(x)          x.begin(), x.end()
#define   gcd(x,y)          __gcd(x,y)
#define   vi              vector<int>
#define   vvi             vector<vector<int>>
#define   mi             map<int,int>
#define   pi              pair<int,int>
#define   vpi             vector<pi>
#define   si              set<int>
#define   MSETI           multiset<int>
#define   inf             1e18
#define   ff              first
#define   ss              second
#define   ps(x,y)         fixed <<setprecision(y)<< x
#define   yes             cout <<"YES" << endl
#define   no              cout <<"NO" << endl
#define   impossible      cout << -1 << endl
#define   endl            "\n"
#define   mp(x,y)         make_pair(x,y)

// ----------------ON A MISSION --------------- //
void pseudo()
{
  go;
#ifndef ONLINE_JUDGE
  freopen("input.txt", "r", stdin);
  freopen("output.txt", "w", stdout);
#endif
}

long long hcf(long long a, long long b)
{
  if (b == 0)  return a;
  a %= b;
  return
    hcf(b, a);
}

long long lcm(long long a, long long b)
{
  return (a * b) / hcf(a, b);
}
//-------------------------------------------------------------------------------------------------//
void solA()
{
  int r, s ;
  cin >> r >> s ;
  int pr = r * s ;
  if (s == 1)
    no ;
  else
  {
    yes ;
    cout << r << " " << pr << " " << r*(s + 1) << endl;
  }

}

void solB()
{
  int n ;
  cin >> n ;
  int arr[n] ;
  readarr(arr, n);
  int temp = 0 , u = 0 , v = 0;

  if (n == 1)
  {
    cout << "0\n" ;
    return ;
  }
  cout << (n + 1) / 2  ;
  cout << endl ;

  for (int  i = 0; i < n; i += 2)
  {
    int mx = min(arr[i], arr[i + 1]) ;
    if (i == n - 1) continue  ;
    cout << i + 1 << " " << i + 2 << " " <<  mod << " " << mx << endl;
    arr[i + 1] =  mx ;
  }
  if (n % 2 == 1)
    cout << n - 1 << " " << n << " " << min(arr[n - 1], arr[n - 2]) << " " << mod << endl;
}



int give( ll  n)
{

  // O(n/2)
  for ( int  k = 0;  k < n - 1;  k += 2)
  {
    int  t ;
    cout << "? 1 " <<  k + 1 << " " <<  k + 2 << " " << n - 1 <<  endl ;

    cin >>  t ;
    if ( t == n) return  k + 1;

    if ( t == n - 1)
    {

      cout << "? 1 " <<  k + 2 << " " <<  k + 1 << " " << n - 1 << endl;
      cin >> t;
      if (t == n)  return  k;
    }
  }
  return n - 1;
}

void solC()
{
  int  n; cin >> n;
  int  arr[n];

  int  myval  = give(n);
  arr[myval] = n;

  f(i, 0, n)
  {
    if (i == myval)   continue;
    cout << "? 2 " << i + 1 << " " << myval + 1 << " " << 1 << endl;
    cin >> arr[i];
  }

  cout << "! ";
  f(i, 0, n)
  {
    cout << arr[i] << " ";
  }
  cout << endl;
}


//-------------------------------------------------------------------------------------------------//
int32_t main()
{
   //pseudo() ;
  w(t)
  {
    solC();
  }

  return 0 ;
}