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 <bits/stdc++.h>
using namespace std;
template <typename T> void read(T &t) {
	t=0; char ch=getchar(); int f=1;
	while (ch<'0'||ch>'9') { if (ch=='-') f=-1; ch=getchar(); }
	do { (t*=10)+=ch-'0'; ch=getchar(); } while ('0'<=ch&&ch<='9'); t*=f;
}
template <typename T> void write(T t) {
	if (t<0) { putchar('-'); write(-t); return; }
	if (t>9) write(t/10);
	putchar('0'+t%10);
}
template <typename T> void writeln(T t) { write(t); puts(""); }
const int maxn=(1e5)+10;
int T,n,AA[maxn],cnt,a[maxn];
int Median(int x,int y,int z) { return x+y+z-max(max(x,y),z)-min(min(x,y),z); }
int query(int x,int y,int z) {
	cnt++;
	//return Median(abs(AA[x]-AA[y]),abs(AA[y]-AA[z]),abs(AA[z]-AA[x]));
	printf("? %d %d %d\n",x,y,z); fflush(stdout);
	scanf("%d",&x); return x;
}
vector<int> g[maxn];
int main() {
	scanf("%d",&T);
	while (T--) {
		scanf("%d",&n);
		//for (int i=1;i<=n;i++) scanf("%d",&AA[i]);
		int x,y,z; srand(time(0));
		while (1) {
			x=rand()%n+1,y=rand()%n+1,z=rand()%n+1;
			if (x==y||y==z||x==z) continue;
			if (query(x,y,z)<=(n-4)/6) break;
		}
		int now,mx=-1,p,p2;
		for (int i=1;i<=n;i++) g[i].clear();
		for (int i=1;i<=n;i++) {
			if (i==x||i==y) continue;
			now=query(x,y,i);
			g[now].push_back(i);
			mx=max(mx,now);
		}
		p=g[mx][0];
		if ((int)g[mx-1].size()>=2) {
			if (query(x,p,g[mx-1][0])>query(x,p,g[mx-1][1])) swap(g[mx-1][0],g[mx-1][1]);
		}
		p2=g[mx-1][0];
		a[p]=1,a[p2]=2;
		for (int i=1;i<=n;i++) if (i!=p&&i!=p2) {
			now=query(p,p2,i);
			a[i]=now+2;
		}
		if (a[1]>a[2]) for (int i=1;i<=n;i++) a[i]=n-a[i]+1;
		printf("! ");
		for (int i=1;i<=n;i++) printf("%d ",a[i]); puts("");
		fflush(stdout);
		//for (int i=1;i<=n;i++) assert(AA[i]==a[i]);
		//printf("cnt=%d\n",cnt);
		scanf("%d",&x);
	}
	
	
	return 0;
}
/*
  0. Enough array size? Enough array size? Enough array size? Integer overflow?
  
  1. Think TWICE, Code ONCE!
  Are there any counterexamples to your algo?
    
  2. Be careful about the BOUNDARIES!
  N=1? P=1? Something about 0?
    
  3. Do not make STUPID MISTAKES!
  Time complexity? Memory usage? Precision error?
*/