Problem: This is an interactive problem.

There is a secret permutation p (1-indexed) of numbers from 1 to n. More formally, for 1 ≤ i ≤ n, 1 ≤ p[i] ≤ n and for 1 ≤ i < j ≤ n, p[i] ≠ p[j]. It is known that p[1]<p[2].

In 1 query, you give 3 distinct integers a,b,c (1 ≤ a,b,c ≤ n), and receive the median of \{|p[a]-p[b]|,|p[b]-p[c]|,|p[a]-p[c]|\}.

In this case, the median is the 2-nd element (1-indexed) of the sequence when sorted in non-decreasing order. The median of \{4,6,2\} is 4 and the median of \{0,123,33\} is 33.

Can you find the secret permutation in not more than 2n+420 queries?

Note: the grader is not adaptive: the permutation is fixed before any queries are made.

Input

The first line of input contains a single integer t (1 ≤ t ≤ 1000) — the number of testcases.

The first line of each testcase consists of a single integer n (20 ≤ n ≤ 100000) — the length of the secret permutation.

It is guaranteed that the sum of n over all test cases does not exceed 100000.

Interaction

For each testcase, you begin the interaction by reading n.

To perform a query, output "? a b c" where a,b,c is the 3 indices you want to use for the query.

Numbers have to satisfy 1 ≤ a,b,c ≤ n and a ≠ b,b ≠ c,a ≠ c.

For each query, you will receive a single integer x: the median of \{|p[a]-p[b]|,|p[b]-p[c]|,|p[a]-p[c]|\}.

In case your query is invalid or you asked more than 2n+420 queries, the interactor will print "−1" and will finish interaction. You will receive Wrong answer verdict. Make sure to exit immediately to avoid getting other verdicts.

When you have determined the secret permutation, output "! p[1] p[2] ... p[n]". If the secret permutation is correct, the interactor will print "1". Otherwise, the interactor will print "-1" and will finish interaction. You will receive Wrong answer verdict. Make sure to exit immediately to avoid getting other verdicts.

After printing a query do not forget to output the end of line and flush the output. Otherwise, you will get Idleness limit exceeded verdict.

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 documentation for other languages.



Hacks:

To hack, use the following format of test:

The first line should contain a single integer t (1 ≤ t ≤ 1000) — the number of testcases.

The first line of each testcase should contain a single integer n (20 ≤ n ≤ 100000) — the length of the secret permutation.

The following line of should contain n integers p[1],p[2],p[3],…,p[n]. p[1]<p[2] and p must be a permutation of integers from 1 to n.

You must ensure that the sum of n over all testcases does not exceed 100000.

Example

Input


1
20

6

9

1

Output




? 1 5 2

? 20 19 2

! 9 10 19 7 16 18 11 14 15 6 20 8 17 4 5 3 12 2 13 1

Note

The secret permutation is \{9,10,19,7,16,18,11,14,15,6,20,8,17,4,5,3,12,2,13,1\}.

For the first query, the values of (a,b,c) is (1,5,2). Since p[1]=9, p[5]=16 and p[2]=10. The return value is the median of \{|9-16|,|16-10|,|9-10|\} which is 6.

For the second query, the values of (a,b,c) is (20,19,2). Since p[20]=1, p[19]=13 and p[2]=10. The return value is the median of \{|1-13|,|13-10|,|1-10|\} which is 9.

By some miracle, we have figured out that the secret permutation is \{9,10,19,7,16,18,11,14,15,6,20,8,17,4,5,3,12,2,13,1\}. We output it and receive 1 from the interactor, meaning that we have guessed the secret permutation correctly.

Solution (in CPP):

Solution:
#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
//#define endl '\n'
#define ll long long
#define pi pair<int, int>
#define f first
#define s second
 
const int mxn = 100000, k = 13;
int n;
int a[mxn];
 
int qry(int x, int y, int z){
	int ret;
	cout << "? " << (x + 1) << " " << (y + 1) << " " << (z + 1) << endl;
	cin >> ret;
	return ret;
}
 
void answer(){
	cin >> n;
	
	int x, y;
	for(x = 0; x < k; x++)
	for(y = 0; y < x; y++)
	for(int i = 0; i < y; i++){
		if(qry(x, y, i) <= (n - 4) / 6) goto hell;
	}
	
	hell:
	int z = 0;
	for(int i = 0; i < n; i++){
		a[i] = i != x && i != y ? qry(x, y, i) : 0;
		if(a[i] > a[z]) z = i;
	} 
	
	int w = 0;
	while(a[w] != a[z] - 1) w++;
	
	for(int i = w + 1; i < n; i++){
		if(a[i] == a[w] && qry(x, z, w) > qry(x, z, i)) w = i;
	} 
	
	a[z] = 0, a[w] = 1;
	for(int i = 0; i < n; i++) if(i != z && i != w) a[i] = qry(w, z, i) + 1;
	
	if(a[0] > a[1]) for(int i = 0; i < n; i++) a[i] = n - a[i] - 1;
	
	cout << "!";
	for(int i = 0; i < n; i++) cout << " " << (a[i] + 1);
	cout << endl;
	
	int ret;
	cin >> ret;
	
	if(!~ret) exit(0);
}
 
int main(){
	ios::sync_with_stdio(0);
	cin.tie(0);
	
	int t;
	cin >> t;
	
	for(int i = 0; i < t; i++) answer();
 
	return 0;
}