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:
// Motto:-Fail early,Fail Often,Fail Forward.
//  while(true)
// {
//     if(AC)
//     {
//         break;
//     }
//     else if(Contest Over)
//     {
//         Try.
//         Check out Editorial.
//         Understand.
//         Find out your Mistake.
//         Learn the topic (if new).
//         Solve Problems on that topic (if new).
//         Upsolve that problem.
//         break;
//     }
//     else
//     {
//         Try.
//         Use Pen-Paper.
//         Find errors, edge cases, etc.
//         continue;
//     }
// }
// Getting TLE ? Do check for long long int optimisation.
    #include<bits/stdc++.h>
    using namespace std;
    //=============================LOL======================
    #define int         long long
    #define double      long double
    #define pb          push_back
    #define pf          push_front
    #define pii         pair<int,int>
    #define vi          vector<int>
    #define vnn         vector<vector<int>>
    #define vii         vector<pii>
    #define vs          vector<string>
    #define all(a)      (a).begin(),(a).end()
    #define rall(a)     (a).rbegin(),(a).rend()
    #define x           first
    #define y           second
    // #define endl        '\n'
    #define sz(x)       (int)(x).size()
    #define thunder     ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0)
    //=============================LOL========================
    //++++++++++++++++++++++++Optimisations+++++++++++++++
    //#pragma GCC optimize("O1") 
    //#pragma GCC optimize("O2")
    //#pragma GCC optimize("O3") 
    //#pragma GCC optimize("Os") 
    //#pragma GCC optimize("Ofast") 
    //#pragma GCC optimize("Ofast")  
    //#pragma GCC target("avx,avx2,fma")
    //++++++++++++++++++++++++++++++++++++++++++++++++++++++ 
    //=========================Chal,Nikal.ab==========================
    #define fo(i,l,u)   for(i=l;i<u;i++)
    #define rfo(i,l,u)  for(i=l;i>=u;i--)
    #define allfo(s)    for(auto it=(s).begin();it!=(s).end();it++)
    #define _init(b)     memset(b,-1,sizeof(b))
    #define _init0(b)     memset(b,0,sizeof(b))
    #define MOD         1000000007
    #define hell        998244353
    #define output(x) cout << (x ? "YES" : "NO")<<endl;
    #define Output(x) cout << (x ? "Yes" : "No")<<endl;
    //=========================Chal,Nikal.ab=============================
    //---------------------------Hushiyari mat dikha---------------------------------
    #define MOD2 (998244353)
    #define MOD3 (1000000009)
    #define PI acos(-1)
    #define eps (1e-8)
    #define INF (1e18) 
    template<class A,class B>ostream&operator<<(ostream&out,const pair<A,B>&a){return out<<"("<<a.first<<","<<a.second<<")";}
    template<class A>ostream&operator<<(ostream&out,const vector<A>&a){for(const A &it:a)out<<it<<" ";return out;}
    template<class A,class B>istream&operator>>(istream&in,pair<A,B>&a){return in>>a.first>>a.second;}
    template<class A>istream&operator>>(istream&in,vector<A>&a){for(A &i:a)in>>i;return in;}
    ifstream cinn("in.txt");ofstream coutt("out.txt");
    int poww(const int &a,int b,const int &m=MOD){if(b==0)return 1;int x=poww(a,b/2,m);x=x*x%m;if(b&1)x=x*a%m;return x;} 
    //---------------------------Hushiyari mat dikha--------------------------------------
    int gcd(int a, int b) 
    { 
    if (b == 0) 
        return a; 
    return gcd(b, a % b);    
    }
    int _sum(int n)
    {
        return (n*(n+1))/2;
    }
    int distSq(pii p, pii q) 
    { 
    return (p.x - q.x) * (p.x - q.x) + (p.y - q.y) * (p.y - q.y);  
    }
    vi fact(101);
    int nCr(int n, int r)
    {
       if(!r) return 1;
       return (n*nCr(n-1,r-1))/r;
    }
    const int N = 2*1e5 + 5;
    // int k,mx=-1,spe;
    vi adj[N],vis(N,0),dfsvis(N,0);
    // int dis[N];
    // map<int,int> mm;
    // bool isPalindrome(string str)
    // {
    // int low = 0;
    // int high = str.length() - 1;
 
    // while (low < high)
    // {
    //     // if a mismatch happens
    //     if (str[low] != str[high]) {
    //         return false;
    //     }
 
    //     low++;
    //     high--;
    // }
 
    // return true;
    // }
    
    // bool comp (string a,string b)
    // {
    //     if(a.length()==b.length())
    //     {
    //         return a<b;
    //     }
    //     return a.length()<b.length();
    // }
    // void build_tree(int l,int r,vi &v,vi &d,int curd)
    // {
    //     if(l>r) return;
    //     if(l==r)
    //     {
    //         d[l]=curd;
    //     }
    //     int mx=l,i;
    //     fo(i,l+1,r+1)
    //     {
    //         if(v[mx]<v[i])
    //         {
    //             mx=i;
    //         }
    //     }
    //     d[mx]=curd;
    //     build_tree(l,mx-1,v,d,curd+1);
    //     build_tree(mx+1,r,v,d,curd+1);

    // }
    // int ok=1;
    // vii p;
    // int mn=1e18;
    // void dfs(int node)
    // {
    //     vis[node]=1;
    //     for(int n : adj[node])
    //     {
    //         if(!vis[n])
    //             dfs(n);
    //     }
    // }
    // int th=-1;
    // void dfs1(int node,int c,int spe)
    // {
    //     vis[node]=1;
    //     if(node==spe) 
    //     {
    //         th=c;
    //         return;
    //     }
    //     // if(sz(adj[node])!=2) ok=0;
    //     for(int n : adj[node])
    //     {
    //         if(!vis[n])
    //         {
    //             dfs1(n,c+1,spe);
    //         }
    //     }
    // }
    // bool cycleBfs(int node)
    // {
    //     queue<pii> q;
    //     q.push({node,-1});
    //     vis[node]=1;
    //     while(!q.empty())
    //     {
    //         int node =q.front().x;
    //         int prev = q.front().y;
    //         q.pop();
    //         for(int n : adj[node])
    //         {
    //             if(vis[n] && prev!=n) return true;
    //             if(!vis[n])
    //             {
    //                 vis[n]=1;
    //                 q.push({n,node});
    //             }
    //         }
    //     }
    //     return false;
    // }
    // bool cycleDfs(int node,int par)
    // {
    //     vis[node]=1;
    //     for(int n : adj[node])
    //     {
    //         if(!vis[n])
    //         {
    //             if(cycleDfs(n,node)) return true;
    //         }
    //         else if(par!=n)
    //         {
    //             return true;
    //         }
    //     }
    //     return false;
    // }
    // int dp[2][N];
    // int maxlen(vi &v,int prev,int o,int i)
    // {
    //     if(i>=sz(v)) return 0;
    //     if(dp[o][i]!=-1) return dp[o][i];
    //     if(o)
    //     {
    //         int mx=1;
    //         if(prev==-1)
    //         {
    //             mx=max(mx,maxlen(v,v[i],o,i+1)+1);
    //             mx=max(mx,maxlen(v,prev,o,i+1));
    //         }
    //         else
    //         {
    //             if(prev<v[i]) 
    //             {
    //                     mx=max(mx,maxlen(v,v[i],o,i+1)+1);
    //                     mx=max(mx,maxlen(v,prev,o-1,i+1));
    //             }
    //             else
    //             {
    //                 mx=max(mx,maxlen(v,prev,o-1,i+1));
    //             }
    //         }
    //         return dp[o][i]=mx;
    //     }
    //     else
    //     {
    //         int mx=-1;
    //         if(prev==-1)
    //         {
    //             mx=max(mx,maxlen(v,v[i],o,i+1)+1);
    //             mx=max(mx,maxlen(v,prev,o,i+1));
    //         }
    //         else 
    //         {
    //             if(prev<v[i]) 
    //                     mx=max(mx,maxlen(v,v[i],o,i+1)+1);
    //             else
    //                     mx=max(mx,0LL);
    //         }
    //         return dp[o][i]=mx;
    //     }
    // }
    // bool Bipartitebfs(int node)
    // {
    //     color[node]=0;
    //     queue<int> q;
    //     q.push(node);
    //     while(!q.empty())
    //     {
    //         node=q.front();
    //         q.pop();
    //         for(int n : adj[node])
    //         {
    //             if(color[n]==-1)
    //             {
    //                 color[n]=(color[node]^1);
    //                 q.push(n);
    //             }
    //             else if(color[n]==color[node])
    //             {
    //                 return false;
    //             }
    //         }
    //     }
    //     return true;
    // }
    // bool Bipartitedfs(int node,int col)
    // {
    //     color[node]=col;
    //     for(int n : adj[node])
    //     {
    //         if(color[n]==-1)
    //         {
    //             if(!Bipartitedfs(n,col^1)) return false;
    //         }
    //         else if(color[n]==color[node])
    //         {
    //             return false;
    //         }
    //     }
    //     return true;
    // }
    // bool cycleDIRdfs(int node)
    // {
    //     vis[node]=1;
    //     dfsvis[node]=1;

    //     for(int n : adj[node])
    //     {
    //         if(!vis[n])
    //         {
    //             if(cycleDIRdfs(n)) return true;
    //         }
    //         else if(dfsvis[n]) return true;
    //     }
    //     dfsvis[node]=0;
    //     return false;
    // }
    stack<int> s;
    void TopoSortDFS(int node)
    {
        vis[node]=1;
        for(int n : adj[node])
        {
            if(!vis[n]) TopoSortDFS(n);
        }
        s.push(node);
    }
    int query(int t,int i,int j,int x1)
    {
        cout<<"?"<<" "<<t<<" "<<i<<" "<<j<<" "<<x1<<endl;
        cout.flush();
        int a;
        cin>>a;
        return a;
    }
    bool isprime(int n)
    {
        int i;
        if(n==1) return false;
        for(i=2;i*i<=n;i++)
        {
            if(n%i==0) return false;
        }
        return true;
    }


    signed main()
    {
        thunder;
        #ifndef ONLINE_JUDGE
           freopen("Input.txt","r",stdin);
           freopen("Output.txt","w",stdout);
        #endif
        int t,c,n,m,b,f,f2,k,t2,t3,i,j,ans,mnt,mxt,j1,j2;
        cin>>t;
        while(t--)
        {
            cin>>n;
            vi v(n,-1);
            //For Max
            f=0;int ind=-1;
            fo(i,0,n-1)
            {
                int val=query(1,i+1,i+2,n-1);
                if(val==n)
                {
                    f=1;
                    ind=i+1;
                }
                if(val==n-1)
                {
                    int val1=query(1,i+2,i+1,n-1);
                    if(val1==n)
                    {
                        ind=i;
                        f=1;
                    }
                }
                i++;
                if(f) break;
            }
            if(!f) ind=n-1;
            v[ind]=n;
            fo(i,0,n)
            {
                if(i!=ind)
                {
                    int val=query(2,i+1,ind+1,1);
                    v[i]=val;
                }
            }
            cout<<"! ";
            cout<<v<<endl;
            cout.flush();
        }
    }