/*
    Nishant Singh
    nishant.ns19@gmail.com
*/
#include <bits/stdc++.h>
#include <functional> 
#include <iostream> 
// ******** Big Integer ********
// #include <boost/multiprecision/cpp_int.hpp>
// using namespace boost::multiprecision;
// #define ll1024  int1024_t
// ******** Big Integer ********
// ******** Using PBDS ********
// #include <ext/pb_ds/assoc_container.hpp> // Common file 
// #include <ext/pb_ds/tree_policy.hpp> 
// using namespace __gnu_pbds; 
// #define ook order_of_key    //count of strictly smaller elements
// #define fbo find_by_order   //find element at an index 
// ******** Using PBDS ********
using namespace std;
#define ll long long
#define ld long double
#define pll pair<long long,long long>
#define pii pair<int,int>
#define vll vector<ll>
#define vec vector
#define setcount(x) __builtin_popcountll(x)
#define f first
#define s second
#define mp make_pair
#define REP(i,n) for(ll i=0;i<(n);i++)
#define FOR(i,a,b) for(ll i=(a);i<=(b);i++)
#define REV(i,a,b) for(ll i=(a);i>=(b);i--)
#define pb push_back
#define in insert
#define init(x,a) memset(x,a,sizeof(x))
#define all(c) c.begin(),c.end()
#define pq priority_queue
#define ffl fflush(stdout)
#define mset multiset
#define out(x) cout<<#x<<" = "<<" "<<x<<"\n";
#define ub upper_bound
#define lb lower_bound 
#define ln "\n"
#define Y "YES"
#define N "NO"
ll const mod=1e9 + 7;
ll const inf=2e18;
// ******** Using PBDS ********
// typedef tree<long long, null_type, less<long long>, rb_tree_tag, 
//              tree_order_statistics_node_update> 
//                 os; 
// ******** Using PBDS ********
inline ll add(ll a,ll b){ 
    return (a%mod + b%mod + mod)%mod; 
}
inline ll mul(ll a,ll b){ 
    return (a%mod * b%mod + mod)%mod; 
}
inline ll sub(ll a,ll b){ 
    return (a%mod - b%mod + mod)%mod; 
}
ll gcd(ll a,ll b){
    if (b == 0) 
        return a; 
    return gcd(b, a % b);  
}
ll modexp(ll x, ll n){
    if(n==0){
        return 1;
    }
    else if(n&1){
        return (modexp((x*x)%mod, n/2)%mod * x%mod)%mod;
    }
    else{
        return (modexp((x*x)%mod, n/2)%mod);
    }
}
ll modinv(ll n){
    return modexp(n,mod-2);
}
// ******** GLOBAL VARIABLES ********
ll n;
// ******** GLOBAL VARIABLES ********
ll findMax(ll i,ll j){
    cout<<"? 1 "<<i<<" "<<j<<" "<<n-1<<endl;
    ll x;
    cin>>x;
    if(x==-1)
        exit(0);
    return x;
}
ll findMin(ll i,ll j){
    cout<<"? 2 "<<i<<" "<<j<<" "<<1<<endl;
    ll x;
    cin>>x;
    if(x==-1)
        exit(0);
    return x;
}
void solve(){
    cin>>n;
    ll a[n+1];
    ll idx=n;
    for(int i=1;i<n;i=i+2){
        ll val=findMax(i,i+1);
        if(val==(n)){
            idx=i+1;
            break;
        }
        else if(val==(n-1)){
            val=findMax(i+1,i);
            if(val==n){
                idx=i;
                break;
            }
        }
    }
    a[idx]=n;
    for(int i=1;i<=n;i++){
        if(i!=idx){
            a[i]=findMin(i,idx);
        }
    }
    cout<<"! ";
    FOR(i,1,n){
        cout<<a[i]<<" ";
    }
    cout<<endl;
}  
int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    ll T=1;
    cin>>T;  
    for(int t=1;t<=T;t++)
    {
        // cout<<"Case #"<<t<<": ";
        solve();
    }
    return 0;
}
### Prompt

Please formulate a Cpp solution to the following 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.

### Response

```cpp
/*
    Nishant Singh
    nishant.ns19@gmail.com
*/
#include <bits/stdc++.h>
#include <functional> 
#include <iostream> 
// ******** Big Integer ********
// #include <boost/multiprecision/cpp_int.hpp>
// using namespace boost::multiprecision;
// #define ll1024  int1024_t
// ******** Big Integer ********
// ******** Using PBDS ********
// #include <ext/pb_ds/assoc_container.hpp> // Common file 
// #include <ext/pb_ds/tree_policy.hpp> 
// using namespace __gnu_pbds; 
// #define ook order_of_key    //count of strictly smaller elements
// #define fbo find_by_order   //find element at an index 
// ******** Using PBDS ********
using namespace std;
#define ll long long
#define ld long double
#define pll pair<long long,long long>
#define pii pair<int,int>
#define vll vector<ll>
#define vec vector
#define setcount(x) __builtin_popcountll(x)
#define f first
#define s second
#define mp make_pair
#define REP(i,n) for(ll i=0;i<(n);i++)
#define FOR(i,a,b) for(ll i=(a);i<=(b);i++)
#define REV(i,a,b) for(ll i=(a);i>=(b);i--)
#define pb push_back
#define in insert
#define init(x,a) memset(x,a,sizeof(x))
#define all(c) c.begin(),c.end()
#define pq priority_queue
#define ffl fflush(stdout)
#define mset multiset
#define out(x) cout<<#x<<" = "<<" "<<x<<"\n";
#define ub upper_bound
#define lb lower_bound 
#define ln "\n"
#define Y "YES"
#define N "NO"
ll const mod=1e9 + 7;
ll const inf=2e18;
// ******** Using PBDS ********
// typedef tree<long long, null_type, less<long long>, rb_tree_tag, 
//              tree_order_statistics_node_update> 
//                 os; 
// ******** Using PBDS ********
inline ll add(ll a,ll b){ 
    return (a%mod + b%mod + mod)%mod; 
}
inline ll mul(ll a,ll b){ 
    return (a%mod * b%mod + mod)%mod; 
}
inline ll sub(ll a,ll b){ 
    return (a%mod - b%mod + mod)%mod; 
}
ll gcd(ll a,ll b){
    if (b == 0) 
        return a; 
    return gcd(b, a % b);  
}
ll modexp(ll x, ll n){
    if(n==0){
        return 1;
    }
    else if(n&1){
        return (modexp((x*x)%mod, n/2)%mod * x%mod)%mod;
    }
    else{
        return (modexp((x*x)%mod, n/2)%mod);
    }
}
ll modinv(ll n){
    return modexp(n,mod-2);
}
// ******** GLOBAL VARIABLES ********
ll n;
// ******** GLOBAL VARIABLES ********
ll findMax(ll i,ll j){
    cout<<"? 1 "<<i<<" "<<j<<" "<<n-1<<endl;
    ll x;
    cin>>x;
    if(x==-1)
        exit(0);
    return x;
}
ll findMin(ll i,ll j){
    cout<<"? 2 "<<i<<" "<<j<<" "<<1<<endl;
    ll x;
    cin>>x;
    if(x==-1)
        exit(0);
    return x;
}
void solve(){
    cin>>n;
    ll a[n+1];
    ll idx=n;
    for(int i=1;i<n;i=i+2){
        ll val=findMax(i,i+1);
        if(val==(n)){
            idx=i+1;
            break;
        }
        else if(val==(n-1)){
            val=findMax(i+1,i);
            if(val==n){
                idx=i;
                break;
            }
        }
    }
    a[idx]=n;
    for(int i=1;i<=n;i++){
        if(i!=idx){
            a[i]=findMin(i,idx);
        }
    }
    cout<<"! ";
    FOR(i,1,n){
        cout<<a[i]<<" ";
    }
    cout<<endl;
}  
int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    ll T=1;
    cin>>T;  
    for(int t=1;t<=T;t++)
    {
        // cout<<"Case #"<<t<<": ";
        solve();
    }
    return 0;
}
```