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"
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
// #pragma GCC optimization ("O3")
// #pragma GCC optimization ("unroll-loops")
// #include <ext/pb_ds/assoc_container.hpp>
// #include <ext/pb_ds/tree_policy.hpp>
// using namespace __gnu_pbds;
//#include <boost/multiprecision/cpp_int.hpp>
//using namespace boost::multiprecision;
using namespace std;

#define all(c) (c).begin(),(c).end()
// #define endl "\n"
#define ff first
#define ss second
#define allr(c) (c).rbegin(),(c).rend()
#define ifr(i, begin, end) for (__typeof(end) i = (begin) - ((begin) > (end)); i != (end) - ((begin) > (end)); i += 1 - 2 * ((begin) > (end)))
#define pof pop_front
#define pob pop_back
#define pb emplace_back
#define pf emplace_front
#define fstm(m,n,r) m.reserve(n);m.max_load_factor(r)
#define mp make_pair
#define mt make_tuple
#define inf LLONG_MAX
#define os tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update>
//order_of_key (k) : Number of items strictly smaller than k .
//find_by_order(k) : K-th element in a set (counting from zero).
const double PI = acos(-1);
typedef complex<double> cd;
typedef long long ll;
ll gcd(){return 0ll;} template<typename T,typename... Args> T gcd(T a,Args... args) { return __gcd(a,(__typeof(a))gcd(args...)); }
typedef map<ll,ll> mll;
typedef map<string,ll> msll;
typedef unordered_map<ll,ll> umap;
typedef vector<ll> vll;
typedef pair<ll,ll> pll;
typedef long double ld;
#define mod 1000000007 

const int N = 1e5 + 2;

int ask(int t, int i, int j, int x) {
    int r;
    if(t == 1) t = 2; else t = 1;
    cout<<"? "<<t<<" "<<i+1<<" "<<j+1<<" "<<x<<endl;
    cin>>r;
    return r;
}

void answer(vector<int> &a) {
    cout<<"! ";
    for(auto &it: a)
        cout<<it<<" ";
    cout<<endl;
}

void Solve() {
    int n;
    cin>>n;
    int x, y;
    vector<int> a(n,0);
    for(int i=0; i+1<n; i+=2) {
        x = ask(1, i, i+1, 1);
        y = ask(2, i, i+1, n-1);

        if(x == 2 && ask(2, i, i+1, 1)==1) {
            a[i+1] = 1;
            if(y == n-1 && ask(1, i, i+1, n-1)==n)
                a[i] = n;
            else
                a[i] = y;
            continue;
        }

        if(y == n-1 && ask(1, i, i+1, n-1) == n) {
            a[i] = n;
            a[i+1] = x;
            continue;
        }

        if(ask(1, i+1, i, x) == x+1) {
            a[i] = x;
            a[i+1] = y;
        } else {
            a[i] = y;
            a[i+1] = x;
        }
    }
    if(n&1) {
        vector<bool> vis(n+1, 0);
        for(int i=0; i<n-1; i++)
            vis[a[i]] = 1;
        for(int i=1; i<n+1; i++) if(!vis[i]) {
            a[n-1] = i;
            break;
        }
    }
    answer(a);
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    // #ifndef ONLINE_JUDGE
    //     freopen("input.txt","r",stdin);
    // #endif
    
    int T;cin>>T;while(T--) 
    {
        Solve();
    }
    return 0;
}