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:
/* author satish */
#include<bits/stdc++.h>
using namespace                        std;
#define int                            long long
#define pi                             pair<int,int>
#define pq                             priority_queue
#define For(it,x)                      for(auto it=(x).begin();it!=(x).end();it++)
#define endl                           '\n'
#define  min_pq(T)                     pq<T,vector<T>,greater<T>>
#define  max_pq(T)                     pq<T,vector<T>>
#define op(mp)                         mp.reserve(1024); mp.max_load_factor(0.25);
#define all(x)                         x.begin(),x.end()
#define rall(x)                        x.rbegin(),x.rend()
#define found(mp,x)                    mp.find(x)!=mp.end()
#define foundp(mp,i,j)                 mp.find({i,j})!=mp.end()
template<typename T>                   T min(T a,T b,T c)                              {return min(a,min(b,c));}
template<typename T>                   T max(T a,T b,T c)                              {return max(a,max(b,c));}
template<typename T>                   T max(T a,T b,T c,T d)                          {return max(a,max(b,max(c,d)));}
template<typename T>                   T min(T a,T b,T c,T d)                          {return min(a,min(b,min(c,d)));}
template<typename t1>                  void fill(vector<t1>&arr,t1 val)                {for(int i=0;i<(int)arr.size();i++)arr[i]=val;}
template<typename t1>                  void fill(vector<vector<t1>>&arr,t1 val)        {for(int i=0;i<(int)arr.size();i++){for(int j=0;j<(int)arr[i].size();j++)     arr[i][j]=val;}}
template<typename T>                   void read(vector<T>&arr,int N)                  {arr.clear(); arr.resize(N);for(int i=0;i<N;i++)cin>>arr[i];}
template<typename T>                   void read(vector<pair<T,T>>&arr,int N)          {arr.clear();arr.resize(N);for(int i=0;i<(int)arr.size();i++)cin>>arr[i].first>>arr[i].second;}
template<typename T>                   void read(vector<vector<T>>&arr,int N,int M)    {arr.clear();arr.resize(N,vector<T>(M));for(int i=0;i<N;i++){for(int j=0;j<M;j++)cin>>arr[i][j];}}
#define n_l '\n'
#define print(...) cout << "[" << #__VA_ARGS__ << "]: "; cout << to_string(__VA_ARGS__) << endl
template <typename T, size_t N> int SIZE(const T (&t)[N]){ return N; } template<typename T> int SIZE(const T &t){ return t.size(); } string to_string(const string s, int x1=0, int x2=1e9){ return '"' + ((x1 < s.size()) ? s.substr(x1, x2-x1+1) : "") + '"'; } string to_string(const char* s) { return to_string((string) s); } string to_string(const bool b) { return (b ? "true" : "false"); } string to_string(const char c){ return string({c}); } template<size_t N> string to_string(const bitset<N> &b, int x1=0, int x2=1e9){ string t = ""; for(int __iii__ = min(x1,SIZE(b)),  __jjj__ = min(x2, SIZE(b)-1); __iii__ <= __jjj__; ++__iii__){ t += b[__iii__] + '0'; } return '"' + t + '"'; } template <typename A, typename... C> string to_string(const A (&v), int x1=0, int x2=1e9, C... coords); int l_v_l_v_l = 0, t_a_b_s = 0; template <typename A, typename B> string to_string(const pair<A, B> &p) { l_v_l_v_l++; string res = "(" + to_string(p.first) + ", " + to_string(p.second) + ")"; l_v_l_v_l--; return res; } template <typename A, typename... C> string to_string(const A (&v), int x1, int x2, C... coords) { int rnk = rank<A>::value; string tab(t_a_b_s, ' '); string res = ""; bool first = true; if(l_v_l_v_l == 0) res += n_l; res += tab + "["; x1 = min(x1, SIZE(v)), x2 = min(x2, SIZE(v)); auto l = begin(v); advance(l, x1); auto r = l; advance(r, (x2-x1) + (x2 < SIZE(v))); for (auto e = l; e != r; e = next(e)) { if (!first) { res += ", "; } first = false; l_v_l_v_l++; if(e != l){ if(rnk > 1) { res += n_l; t_a_b_s = l_v_l_v_l; }; } else{ t_a_b_s = 0; } res += to_string(*e, coords...); l_v_l_v_l--; } res += "]"; if(l_v_l_v_l == 0) res += n_l; return res; } void dbgm(){;} template<typename Heads, typename... Tails> void dbgm(Heads H, Tails... T){ cout << to_string(H) << " | "; dbgm(T...); }
#define printm(...) cout << "[" << #__VA_ARGS__ << "]: "; dbgm(__VA_ARGS__); cout << endl


static bool comp(pi&a,pi&b){
}

const int Max=2*1e17 , Min=-2*1e17;

const int mod=1e9+7;
int N,M,K;
/*********************************/

int query(int t,int i,int j,int x){
    cout<<"?"<<' '<<t<<' '<<i<<' '<<j<<' '<<x<<endl;
    cout.flush();
    int ret;cin>>ret;
    return ret;
}

int findn(){

    for(int i=1;i<N;i+=2){
        int q=query(1,i,i+1,N-1);
        if(q==N)
            return i+1;
        else if(q==N-1){
            q=query(1,i+1,i,N-1);
            if(q==N)
                return i;
        }
    }
    return N;
}

void solve(){
    cin>>N;
    vector<int>ans(N+1);

    int nind=findn();
    ans[nind]=N;
    for(int i=1;i<=N;i++){
        if(i!=nind){
            ans[i]=query(2,i,nind,1);
        }
    }
    cout<<"! ";
    for(int i=1;i<=N;i++)
        cout<<ans[i]<<' ';
    cout<<endl;
    cout.flush();
}

signed main(){

    bool testcases = 1                ;
    ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);

    if(testcases)                     {int tt; cin>>tt; while(tt--) solve();}
    else                               solve();
}