#include <bits/stdc++.h>
using namespace std;
#define ll long long
const int maxn = 100005;
int T, n, p[maxn], flag = 0;
int ask(int a, int b, int c) {
    assert(1 <= a && a <= n && 1 <= b && b <= n && 1 <= c && c <= n);
    assert(a != b && a != c && b != c);
    cout << "? " << a << " " << b << " " << c << endl;
    if(flag) {
        int x = 0;
        scanf("%d", &x);
        return x;
    }
    int t[3] = {p[a], p[b], p[c]};
    sort(t, t + 3);
    return max(t[1] - t[0], t[2] - t[1]);
}
int ans[maxn];
inline int Rand(int l, int r) {
    return rand() % (r - l + 1) + l;
}
vector<int> A[maxn];
void solve() {
    scanf("%d", &n);
    if(!flag) {
        for(int i = 1; i <= n; i++)
            scanf("%d", &p[i]);
    }
    srand(time(0));
    int a = 0, b = 0, c = 0;
    do {
        do {
            a = Rand(1, n), b = Rand(1, n), c = Rand(1, n);
        }while(a == b || b == c || a == c);
    }while(ask(a, b, c) > (n - 4) / 6);
    for(int i = 1; i <= n; i++)
        A[i].clear();
    for(int i = 1; i <= n; i++)
        if(i != a && i != b) A[ask(a, b, i)].push_back(i);
    int d = n;
    for(; !A[d].size(); d--);
    int pos1 = A[d][0];
    ans[pos1] = 1;
    int pos2 = A[d - 1][0];
    if(A[d - 1].size() >= 2 && ask(pos1, A[d - 1][0], a) > ask(pos1, A[d - 1][1], a)) 
        pos2 = A[d - 1][1];
    ans[pos2] = 2;
    for(int i = 1; i <= n; i++)
        if(i != pos1 && i != pos2)
            ans[i] = ask(pos1, pos2, i) + 2;
    if(ans[1] > ans[2]) {
        for(int i = 1; i <= n; i++)
            ans[i] = n - ans[i] + 1;
    }
    printf("! ");
    for(int i = 1; i <= n; i++)
        printf("%d ", ans[i]);
    cout << endl;
    cin >> n;
}
int main() {
    flag = 1;
    int T = 1;
    if(flag)
        scanf("%d", &T);
    while(T--) solve();
    return 0;
}
### Prompt

In cpp, your task is to solve the following 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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
#define ll long long
const int maxn = 100005;
int T, n, p[maxn], flag = 0;
int ask(int a, int b, int c) {
    assert(1 <= a && a <= n && 1 <= b && b <= n && 1 <= c && c <= n);
    assert(a != b && a != c && b != c);
    cout << "? " << a << " " << b << " " << c << endl;
    if(flag) {
        int x = 0;
        scanf("%d", &x);
        return x;
    }
    int t[3] = {p[a], p[b], p[c]};
    sort(t, t + 3);
    return max(t[1] - t[0], t[2] - t[1]);
}
int ans[maxn];
inline int Rand(int l, int r) {
    return rand() % (r - l + 1) + l;
}
vector<int> A[maxn];
void solve() {
    scanf("%d", &n);
    if(!flag) {
        for(int i = 1; i <= n; i++)
            scanf("%d", &p[i]);
    }
    srand(time(0));
    int a = 0, b = 0, c = 0;
    do {
        do {
            a = Rand(1, n), b = Rand(1, n), c = Rand(1, n);
        }while(a == b || b == c || a == c);
    }while(ask(a, b, c) > (n - 4) / 6);
    for(int i = 1; i <= n; i++)
        A[i].clear();
    for(int i = 1; i <= n; i++)
        if(i != a && i != b) A[ask(a, b, i)].push_back(i);
    int d = n;
    for(; !A[d].size(); d--);
    int pos1 = A[d][0];
    ans[pos1] = 1;
    int pos2 = A[d - 1][0];
    if(A[d - 1].size() >= 2 && ask(pos1, A[d - 1][0], a) > ask(pos1, A[d - 1][1], a)) 
        pos2 = A[d - 1][1];
    ans[pos2] = 2;
    for(int i = 1; i <= n; i++)
        if(i != pos1 && i != pos2)
            ans[i] = ask(pos1, pos2, i) + 2;
    if(ans[1] > ans[2]) {
        for(int i = 1; i <= n; i++)
            ans[i] = n - ans[i] + 1;
    }
    printf("! ");
    for(int i = 1; i <= n; i++)
        printf("%d ", ans[i]);
    cout << endl;
    cin >> n;
}
int main() {
    flag = 1;
    int T = 1;
    if(flag)
        scanf("%d", &T);
    while(T--) solve();
    return 0;
}
```