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>

const int N = 100005;

int query(int a, int b, int c) {
  printf("? %d %d %d\n", a, b, c);
  fflush(stdout);
  int ans;
  return scanf("%d", &ans), ans;
}

int ans[N];

void solve(int n) {
  int a = 0, b = 0;
  for (int x = 1; x <= 13 && !a; ++x) {
    for (int y = x + 1; y <= 13 && !a; ++y) {
      for (int z = y + 1; z <= 13 && !a; ++z) {
        if (query(x, y, z) <= (n - 4) / 6) a = x, b = y;
      }
    }
  }
  std::map<int, std::vector<int>> mp;
  for (int i = 1; i <= n; ++i) {
    if (i != a && i != b) mp[query(a, b, i)].push_back(i);
  }
  int v = mp.rbegin()->first;
  int p1 = mp[v].front(), x1 = mp[v - 1][0];
  if (mp[v - 1].size() > 1) {
    int x2 = mp[v - 1][1];
    if (query(p1, x1, a) > query(p1, x2, a)) std::swap(x1, x2);
  }
  ans[p1] = 1, ans[x1] = 2;
  for (int i = 1; i <= n; ++i) {
    if (i != p1 && i != x1) ans[i] = query(p1, x1, i) + 2;
  }
  if (ans[1] > ans[2]) {
    for (int i = 1; i <= n; ++i) ans[i] = n - ans[i] + 1;
  }
  putchar('!');
  for (int i = 1; i <= n; ++i) printf(" %d", ans[i]);
  puts("");
  fflush(stdout);
  int sta;
  scanf("%d", &sta);
}

int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    int n;
    scanf("%d", &n);
    solve(n);
  }
  return 0;
}