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:
#include <bits/stdc++.h>
#include<unordered_map>
#include <string> 
using namespace std;
#define ll long long
#define int long long
#define lmin -9223372036854775808
#define lmax 9223372036854775807
#define imax INT_MAX
#define imin INT_MIN
#define ffor(i,a,b) for(int i=a;i<b;i++)
#define bfor(i,a,b) for(int i=a-1;i>=b;i--)
#define pb push_back
#define mset(f,a) memset(f,a,sizeof(f))
#define MAXN   10000001
#define mp make_pair
#define double long double
#define pii(x,y) pair<x,y>
#define min3(a,b,c) min(a,min(b,c))
#define max3(a,b,c) max(a,max(b,c))
#define all(x) x.begin(),x.end()
#define allr(x) x.rbegin(),x.rend()
#define sz(x) (ll)x.size()
#define ff first
#define ss second
#define lb lower_bound
#define ub upper_bound
#define bs binary_search
#define vec(x) vector<x > 
#define set(x) set<x > 
#define mulset(x) multiset<x >
#define umap(x,y) unordered_map<x,y>
#define map(x,y) map<x,y>
#define bb begin()
#define ee end()
#define fastio ios_base::sync_with_stdio(false); cin.tie(NULL);

ll M=1e9+7;


ll max(ll x,ll y)
{
    if(x>y)
        return x;
    
    return y;
}


ll min(ll x, ll y)
{
    if(x<y)
        return x;
    
    return y;
}

ll power(ll x ,ll y)
{
    if(y==0)
        return 1;
    
    ll t=power(x, y/2);
    
    t=((t%M)*(t%M))%M;
    
    if(y%2==1)
        t=((t%M)*(x%M))%M;
    
    return t;
    
}
 
// ll getValUtil(ll* st,ll l, ll r, ll lq, ll rq, ll i)
// {
//     if(r<lq || l>rq)
//         return 0;
    
//     if(lq<=l && r<=rq)
//         return st[i];
    
    
//     ll mid=(l+r)/2;
 
//     ll d55=getValUtil(st,l,mid,lq,rq,2*i+1);
//     ll d2=getValUtil(st,mid+1,r,lq,rq,2*i+2);


//     return d55+d2;
// }
 
// ll getVal(ll n, ll *st,ll lq, ll rq)
// {
//     if(lq>rq)
//         return lmax;
    
//     return getValUtil(st,0,n-1,lq,rq,0);
// }
 
// ll updateUtil(ll*st, ll l ,ll r, ll ind, ll i, ll x)
// {
 
//     if(l>ind || r<ind)
//         return st[i];
    
//     if(l==r)
//     {
//         st[i]+=x;
//         return st[i];
//     }
 
//     ll mid=(l+r)/2;
 
//     ll d1=updateUtil(st,l,mid,ind,2*i+1,x);
//     ll d2=updateUtil(st,mid+1,r,ind,2*i+2,x);
 
//     st[i]=st[2*i+1]+st[2*i+2];
    
//     return st[i];
 
// }
 
// ll update(ll n, ll *st, ll ind,ll x)
// {
//     return updateUtil(st,0,n-1,ind,0,x);
// }
 
// void constructUtil(ll*st, ll l, ll r, ll i)
// {
//     if(l==r)
//     {
//         st[i]=0;
//         return;
//     }
 
//     ll mid=(l+r)/2;
 
//     constructUtil(st,l,mid,2*i+1);
//     constructUtil(st,mid+1,r,2*i+2);
 
//     st[i]=0;
 
//     return;
// }
 
// ll* construct(ll n)
// {
//     ll d=log2(n)+1;
//     ll x=power(2,d+1); 
 
//     ll *st=new ll[x+1];
 
//     constructUtil(st,0,n-1,0);
 
//     return st;
// }

void output()
{
    fastio
    
    ll n;
    cin>>n;

    // cin>>n;

    ll i;
    ll ind=-1;

    for(i=0;i<n;i+=2)
    {
        if(i+1>=n)
            break;
        
        cout<<"? 2 "<<i+1<<" "<<i+2<<" "<<1<<endl;
        ll x;
        cin>>x;

        if(x==1)
        {
            ind=i;
            break;
        }
        
        if(x==2)
        {
            cout<<"? 2 "<<i+2<<" "<<i+1<<" "<<1<<endl;
            cin>>x;

            if(x==1)
            {
                ind=i+1;
                break;
            }
        }
    }

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

    ll a[n];
    a[ind]=1;

    for(i=0;i<n;i++)
    {
        if(i==ind)
            continue;
        
        cout<<"? 1 "<<ind+1<<" "<<i+1<<" "<<n-1<<endl;
        ll b;
        cin>>b;

        a[i]=b;
    }

    cout<<"! ";
    for(i=0;i<n;i++)
        cout<<a[i]<<" ";
    cout<<endl;


}
signed main()
{
    #ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    // for writing output to output.txt
    freopen("output.txt", "w", stdout);
    #endif  

    fastio

    
    ll i;

    // fact[0]=1;

    // for(i=1;i<=400000;i++)
    //     fact[i]=(fact[i-1]*i)%M;
    
    ll t;
    t=1;
    cin>>t;

   
    while(t--)
        output();
}