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 <iostream>
#include <string>
#include <vector>
#include <algorithm>
#include <sstream>
#include <queue>
#include <deque>
#include <bitset>
#include <iterator>
#include <list>
#include <stack>
#include <map>
#include <set>
#include <functional>
#include <numeric>
#include <utility>
#include <limits>
#include <time.h>
#include <math.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <assert.h>
#include <cstdio>
#include <cassert>

using namespace std;

// #define scd(t) scanf("%d",&t)
// #define scld(t) scanf("%ld",&t)
// #define sclld(t) scanf("%lld",&t)
// #define scc(t) scanf("%c",&t)
// #define scs(t) scanf("%s",&t)
// #define scf(t) scanf("%f",&t)
// #define sclf(t) scanf("%lf",&t)
#define FIO ios_base::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);
#define FOR(i,j,k,n) for(int i=j;i<=k;i += n)
#define RFOR(i,j,k,n) for(int i=j;i>=k;i -= n)
#define REP(i,j) FOR(i,0,j,1)
#define RREP(i,j) RFOR(i,j,0,1)
#define all(cont) (cont).begin(), (cont).end()
#define rall(cont) (cont).rbegin(), (cont).rend()
#define FOREACH(it,l) for(auto it = l.begin();it != l.end();it++)
#define IN(A,B,C) assert(B <= A && A <= C)
#define MEM(a,b) memset(a,(b),sizeof(a))
#define ff first
#define ss second
#define pb push_back
#define mp make_pair
#define endl "\n"
#define space " "
#define interact fflush(stdout)
#define present(cont,x) ((cont).find(x) != (cont).end())
#define ipresent(cont,x) (find(all(cont),x) != cont.end())
#define sz(cont) int((cont).size())
#define mem(cont) memset(cont,0,sizeof(cont))
#define pi 3.1415926535897932384626433832795

const int LOGN = 20;
const int MOD = 1e9 + 7;
const int MAXN = 2e5 + 5;
const int MAXA = 1e7 + 5;
const int inf = 1e9 + 1;
const int eps = 1e-9;
const long long int inf2 = 1e18 + 1;

typedef double ld;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<string> vs;
typedef vector<pii> vii;
typedef vector<vi> vvi;
typedef vector<vl> vvl;
typedef map<int,int> mpii;
typedef map<ll,ll> mpll;
// typedef unordered_map<int,int> umpii;
// typedef unordered_map<ll,ll> umpll;
typedef set<int> seti;
typedef set<ll> setl;
typedef multiset<int> mseti;
typedef multiset<ll> msetl;

// ll fact[405];		// factorial vector wrt mod
// ll inv[405];		// inverse factorial vector wrt mod
// ll choose[405][405]; // ncr wrt mod
// ll pow2[405];		// power of 2 wrt mod
// setl square;
// vl primes;
// ll A[MAXA];

// void allSquares()
// {
// 		for(int i = 1;i*i<=MAXN;i++)
// 			square.insert(i*i);
// 		return;
// }

// void FactorsSieve()
// {
// 	for(ll i = 0;i<MAXA;i++)
// 		A[i] = i;
// 	for(int i=2;i<MAXA;i++)
// 	{
// 		if(A[i] == i)
// 		{
// 			primes.pb(i);
// 			for(int j=2*i;j<MAXA;j += i)
// 				A[j] = i;
// 		}
// 	}
// 	return;
// }

// vl factorize(ll x)
// {
// 	vl factors;
// 	while(x > 1)
// 	{
// 		factors.pb(A[x]);
// 		x /= A[x];
// 	}
// 	return factors;
// }

// bool comparesecond(const pair<int,int> &a,const pair<int,int> &b)		// for comparing wrt to ss
// {
//     return (a.second < b.second);
// }

// ll fastexp(ll a,ll b,ll md)
// {
// 	ll res = 1;
// 	while(b)
// 	{
// 		if(b & 1)
// 			res = res*a%md;
// 		a = a*a%md;
// 		b >>= 1;
// 	}
// 	return res;
// }

// void Factorial()		// factorial function
// {
// 	fact[0] = 1;
// 	inv[0] = 1;
// 	for(int i=1;i<=n;i++)
// 	{
// 		fact[i] = (i*fact[i-1])%modulo;
// 		inv[i] = fastexp(fact[i],modulo - 2);
// 	}
// 	for(int i=0;i<=n;i++)
// 	{
// 		for(int j=0;j<=i;j++)
// 		{
// 			choose[i][j] = ((fact[i]*inv[j])%modulo * inv[i-j])%modulo;
// 		}
// 	}
// 	for(int i=0;i<=n;i++)
// 		pow2[i] = fastexp(2,i);
// 	return;
// }

int ask(int t,int i,int j,int x)
{
	cout << "? " << t << space << i + 1 << space << j + 1 << space << x << endl;
	int val;
	cin >> val;
	if(val == -1)
		exit(0);
	return val;
}

void solve()
{
	int n;
	cin >> n;
	vi p(n,-1);
	FOR(i,1,n-1,2)
	{
		int pos1 = i - 1;
		int pos2 = i;
		int val = ask(1,pos1,pos2,n-1);
		if(val == n - 1)
		{
			val = ask(1,pos2,pos1,n-1);
			if(val == n)
			{
				p[pos1] = val;
				p[pos2] = ask(2,pos2,pos1,1);
				continue;
			}
		}
		int get = ask(1,pos1,pos2,val-1);
		if(get == val)
		{
			p[pos2] = val;
			p[pos1] = ask(2,pos1,pos2,1);
		}
		if(get == val - 1)
		{
			p[pos1] = val;
			p[pos2] = ask(2,pos2,pos1,1);
		}
	}
	if(p.back() == -1)
	{
		vector<bool> us(n+1);
		REP(i,n-2)
			us[p[i]] = true;
		FOR(i,1,n,1)
		{
			if(!us[i])
				p[p.size() - 1] = i;
		}
	}
	cout << "! ";
	REP(i,n-1)
		cout << p[i] << space;
	cout << endl;
	return;
}

int main()
{
	// FIO;
	// freopen("input.txt","r",stdin);
	// freopen("output.txt","w",stdout);
	// FactorsSieve();
	// Factorial();
	// allSquares();
	// primefactors();
	ll cases = 1;
	cin >> cases;
	for(ll test = 1;test <= cases;test++)
	{
		// cout << "Case #" << test << ": ";
		solve();
	}
	return 0;
}