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>
#define all(v) v.begin(), v.end()
#define rall(v) v.rbegin(), v.rend()
#define rep(i,n) for(int i=0;i<(int)(n);i++)
#define drep(i,j,n) for(int i=0;i<(int)(n-1);i++)for(int j=i+1;j<(int)(n);j++)
#define trep(i,j,k,n) for(int i=0;i<(int)(n-2);i++)for(int j=i+1;j<(int)(n-1);j++)for(int k=j+1;k<(int)(n);k++)
#define codefor int test;scanf("%d",&test);while(test--)
#define INT(...) int __VA_ARGS__;in(__VA_ARGS__)
#define LL(...) ll __VA_ARGS__;in(__VA_ARGS__)
#define yes(ans) if(ans)printf("yes\n");else printf("no\n")
#define Yes(ans) if(ans)printf("Yes\n");else printf("No\n")
#define YES(ans) if(ans)printf("YES\n");else printf("NO\n")
#define popcount(v) __builtin_popcount(v)
#define vector1d(type,name,...) vector<type>name(__VA_ARGS__)
#define vector2d(type,name,h,...) vector<vector<type>>name(h,vector<type>(__VA_ARGS__))
#define vector3d(type,name,h,w,...) vector<vector<vector<type>>>name(h,vector<vector<type>>(w,vector<type>(__VA_ARGS__)))
#define umap unordered_map
#define uset unordered_set
using namespace std;
using ll = long long;
const int MOD=1000000007;
const int MOD2=998244353;
const int INF=1<<30;
const ll INF2=(ll)1<<60;
//入力系
void scan(int& a){scanf("%d",&a);}
void scan(long long& a){scanf("%lld",&a);}
template<class T,class L>void scan(pair<T, L>& p){scan(p.first);scan(p.second);}
template<class T> void scan(T& a){cin>>a;}
template<class T> void scan(vector<T>& vec){for(auto&& it:vec)scan(it);}
void in(){}
template <class Head, class... Tail> void in(Head& head, Tail&... tail){scan(head);in(tail...);}
//出力系
void print(const int& a){printf("%d",a);}
void print(const long long& a){printf("%lld",a);}
void print(const double& a){printf("%.15lf",a);}
template<class T,class L>void print(const pair<T, L>& p){print(p.first);putchar(' ');print(p.second);}
template<class T> void print(const T& a){cout<<a;}
template<class T> void print(const vector<T>& vec){if(vec.empty())return;print(vec[0]);for(auto it=vec.begin();++it!= vec.end();){putchar(' ');print(*it);}}
void out(){putchar('\n');}
template<class T> void out(const T& t){print(t);putchar('\n');}
template <class Head, class... Tail> void out(const Head& head,const Tail&... tail){print(head);putchar(' ');out(tail...);}
//デバッグ系
template<class T> void dprint(const T& a){cerr<<a;}
template<class T> void dprint(const vector<T>& vec){if(vec.empty())return;cerr<<vec[0];for(auto it=vec.begin();++it!= vec.end();){cerr<<" "<<*it;}}
void debug(){cerr<<endl;}
template<class T> void debug(const T& t){dprint(t);cerr<<endl;}
template <class Head, class... Tail> void debug(const Head& head, const Tail&... tail){dprint(head);cerr<<" ";debug(tail...);}
ll intpow(ll a, ll b){ ll ans = 1; while(b){ if(b & 1) ans *= a; a *= a; b /= 2; } return ans; }
ll modpow(ll a, ll b, ll p){ ll ans = 1; while(b){ if(b & 1) (ans *= a) %= p; (a *= a) %= p; b /= 2; } return ans; }
ll updivide(ll a,ll b){if(a%b==0) return a/b;else return (a/b)+1;}
template<class T> void chmax(T &a,const T b){if(b>a)a=b;}
template<class T> void chmin(T &a,const T b){if(b<a)a=b;}

int ask(int t,int i,int j,int x){
    int ret;
    cout<<"? "<<t<<" "<<i+1<<" "<<j+1<<" "<<x<<endl;
    cin>>ret;
    return ret;
}

int main(){
    codefor{
        INT(n);
        vector<int> a(n,-1);
        int v,mi=-1;
        if(n%2==0){
            for(int i=0;i<n;i+=2){
                v=ask(1,i,i+1,n-1);
                if(v==n){
                    a[i+1]=n;
                    mi=i+1;
                    break;
                }else if(v==n-1){
                    v=ask(1,i+1,i,n-1);
                    if(v==n){
                        a[i]=n;
                        mi=i;
                        break;
                    }
                }
            }
        }else{
            for(int i=0;i<n-1;i+=2){
                v=ask(1,i,i+1,n-1);
                if(v==n){
                    a[i+1]=n;
                    mi=i+1;
                    break;
                }else if(v==n-1){
                    v=ask(1,i+1,i,n-1);
                    if(v==n){
                        a[i]=n;
                        mi=i;
                        break;
                    }
                }
            }
            if(mi==-1){
                a[n-1]=n;
                mi=n-1;
            }
        }
        assert(mi!=-1);
        for(int i=0;i<n;i++){
            if(a[i]==-1){
                v=ask(2,i,mi,1);
                a[i]=v;
            }
        }
        cout<<"!";
        rep(i,n)cout<<" "<<a[i];
        cout<<endl;
    }
}