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<cstdio>
#include<algorithm>
#include<cmath>
using namespace std;
const int maxn=1e5;
int T,n;
int val[maxn+5];
int p[maxn+5];
int query(int a,int b,int c){
	int x;
	printf("? %d %d %d\n",a,b,c);
	fflush(stdout);
	scanf("%d",&x);
	return x;
}
int main(){
	scanf("%d",&T);
	while(T--){
		int a,b;
		scanf("%d",&n);
		bool ok=0;
		for(int i=1;i<=13&&!ok;i++)
			for(int j=i+1;j<=13&&!ok;j++)
				for(int k=j+1;k<=13&&!ok;k++)
					if(query(i,j,k)*6<=n-4){
						a=i; b=j; ok=1;
					}
		int nowid,nows=0;
		int y[5],toty=0;
		for(int i=1;i<=n;i++){
			val[i]=0;
			if(i==a||i==b) continue;
			val[i]=query(a,b,i);
			if(val[i]>nows){
				nows=val[i]; nowid=i;
			}
		}
		for(int i=1;i<=n;i++)
			if(val[i]==nows-1) y[++toty]=i;
		if(toty==1){ a=nowid; b=y[1]; }
		else{
			int v1=query(nowid,a,y[1]),v2=query(nowid,a,y[2]);
			a=nowid;
			if(v1<v2) b=y[1];
			else b=y[2];
		}
		p[a]=1; p[b]=2;
		for(int i=1;i<=n;i++)
			if(i!=a&&i!=b) p[i]=query(a,b,i)+2;
		if(p[1]>p[2])
			for(int i=1;i<=n;i++) p[i]=n-p[i]+1;
		printf("! ");
		for(int i=1;i<=n;i++) printf("%d ",p[i]);
		printf("\n");
		fflush(stdout);
		scanf("%d",&n);
	}
	return 0;
}