It turns out that the meaning of life is a permutation p_1, p_2, …, p_n of the integers 1, 2, …, n (2 ≤ n ≤ 100). Omkar, having created all life, knows this permutation, and will allow you to figure it out using some queries.

A query consists of an array a_1, a_2, …, a_n of integers between 1 and n. a is not required to be a permutation. Omkar will first compute the pairwise sum of a and p, meaning that he will compute an array s where s_j = p_j + a_j for all j = 1, 2, …, n. Then, he will find the smallest index k such that s_k occurs more than once in s, and answer with k. If there is no such index k, then he will answer with 0.

You can perform at most 2n queries. Figure out the meaning of life p.

Interaction

Start the interaction by reading single integer n (2 ≤ n ≤ 100) — the length of the permutation p.

You can then make queries. A query consists of a single line "? \enspace a_1 \enspace a_2 \enspace … \enspace a_n" (1 ≤ a_j ≤ n).

The answer to each query will be a single integer k as described above (0 ≤ k ≤ n).

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



To output your answer, print a single line "! \enspace p_1 \enspace p_2 \enspace … \enspace p_n" then terminate.

You can make at most 2n queries. Outputting the answer does not count as a query.

Hack Format

To hack, first output a line containing n (2 ≤ n ≤ 100), then output another line containing the hidden permutation p_1, p_2, …, p_n of numbers from 1 to n.

Example

Input


5

2

0

1


Output



? 4 4 2 3 2

? 3 5 1 5 5

? 5 2 4 3 1

! 3 2 1 5 4

Note

In the sample, the hidden permutation p is [3, 2, 1, 5, 4]. Three queries were made.

The first query is a = [4, 4, 2, 3, 2]. This yields s = [3 + 4, 2 + 4, 1 + 2, 5 + 3, 4 + 2] = [7, 6, 3, 8, 6]. 6 is the only number that appears more than once, and it appears first at index 2, making the answer to the query 2.

The second query is a = [3, 5, 1, 5, 5]. This yields s = [3 + 3, 2 + 5, 1 + 1, 5 + 5, 4 + 5] = [6, 7, 2, 10, 9]. There are no numbers that appear more than once here, so the answer to the query is 0.

The third query is a = [5, 2, 4, 3, 1]. This yields s = [3 + 5, 2 + 2, 1 + 4, 5 + 3, 4 + 1] = [8, 4, 5, 8, 5]. 5 and 8 both occur more than once here. 5 first appears at index 3, while 8 first appears at index 1, and 1 < 3, making the answer to the query 1.

Note that the sample is only meant to provide an example of how the interaction works; it is not guaranteed that the above queries represent a correct strategy with which to determine the answer.
//package currentContest;
 
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashSet;
import java.util.Random;
import java.util.StringTokenizer;

public class P4 {
	static int N = 1000001;
 
	// Array to store inverse of 1 to N
	static long[] factorialNumInverse = new long[N + 1];
 
	// Array to precompute inverse of 1! to N!
	static long[] naturalNumInverse = new long[N + 1];
 
	// Array to store factorial of first N numbers
	static long[] fact = new long[N + 1];
 
	// Function to precompute inverse of numbers
	public static void InverseofNumber(int p) {
		naturalNumInverse[0] = naturalNumInverse[1] = 1;
 
		for (int i = 2; i <= N; i++)
			naturalNumInverse[i] = naturalNumInverse[p % i] * (long) (p - p / i) % p;
	}
 
	// Function to precompute inverse of factorials
	public static void InverseofFactorial(int p) {
		factorialNumInverse[0] = factorialNumInverse[1] = 1;
 
		// Precompute inverse of natural numbers
		for (int i = 2; i <= N; i++)
			factorialNumInverse[i] = (naturalNumInverse[i] * factorialNumInverse[i - 1]) % p;
	}
 
	// Function to calculate factorial of 1 to N
	public static void factorial(int p) {
		fact[0] = 1;
 
		// Precompute factorials
		for (int i = 1; i <= N; i++) {
			fact[i] = (fact[i - 1] * (long) i) % p;
		}
	}
 
	// Function to return nCr % p in O(1) time
	public static long Binomial(int N, int R, int p) {
 
		// n C r = n!*inverse(r!)*inverse((n-r)!)
		long ans = ((fact[N] * factorialNumInverse[R]) % p * factorialNumInverse[N - R]) % p;
 
		return ans;
	}
 
	public static void main(String[] args) {
		// TODO Auto-generated method stub
		FastReader sc = new FastReader();
//		int p = 998244353;
//	    InverseofNumber(p);
//	    InverseofFactorial(p);
//	    factorial(p);
		// System.out.println(3|1);
		int n=sc.nextInt();
		int ans[]=new int[n];
		Arrays.fill(ans, 0);
		//lets find the last one
		for(int i=2;i<=n;i++) {
			StringBuilder query=new StringBuilder("? ");
			for(int j=0;j<n-1;j++) {
				query.append("1 ");
			}
			query.append(i+" ");
			System.out.println(query);
			System.out.flush();
			int x=sc.nextInt();
			if(x==0) {
				ans[n-1]=n-i+2;
				break;
			}
		}
		if(ans[n-1]==0) {
			ans[n-1]=1;
		}
		for(int i=1;i<=n;i++) {
			if(i==ans[n-1]) {
				continue;
			}
			if(i<ans[n-1]) {
				int tofind=(ans[n-1]-i+1);
				StringBuilder query=new StringBuilder("? ");
				for(int j=0;j<n-1;j++) {
					query.append(tofind+" ");
				}
				query.append(1+" ");
				System.out.println(query);
				System.out.flush();
			}else {
				int tofind=i-ans[n-1]+1;
				StringBuilder query=new StringBuilder("? ");
				for(int j=0;j<n-1;j++) {
					query.append(1+" ");
				}
				query.append(tofind+" ");
				System.out.println(query);
				System.out.flush();
			}
			int x=sc.nextInt();
			ans[x-1]=i;
		}
		StringBuilder finalans=new StringBuilder("! ");
		for(int i=0;i<n;i++) {
			finalans.append(ans[i]+" ");
		}
		System.out.println(finalans);
		System.out.flush();
	}
	public static long gcd1(long a, long b) {
		if (a == 0) {
			return b;
		}
		return gcd1(b % a, a);
	}

	static FastReader sc = new FastReader();
 
	public static void solvegraph() {
		int n = sc.nextInt();
 
		int edge[][] = new int[n - 1][2];
		for (int i = 0; i < n - 1; i++) {
			edge[i][0] = sc.nextInt() - 1;
			edge[i][1] = sc.nextInt() - 1;
		}
		ArrayList<ArrayList<Integer>> ad = new ArrayList<>();
		for (int i = 0; i < n; i++) {
			ad.add(new ArrayList<Integer>());
		}
		for (int i = 0; i < n - 1; i++) {
			ad.get(edge[i][0]).add(edge[i][1]);
			ad.get(edge[i][1]).add(edge[i][0]);
		}
		int parent[] = new int[n];
		Arrays.fill(parent, -1);
		parent[0] = n;
		ArrayDeque<Integer> queue = new ArrayDeque<>();
		queue.add(0);
		int child[] = new int[n];
		Arrays.fill(child, 0);
		ArrayList<Integer> lv = new ArrayList<Integer>();
		while (!queue.isEmpty()) {
			int toget = queue.getFirst();
			queue.removeFirst();
			child[toget] = ad.get(toget).size() - 1;
			for (int i = 0; i < ad.get(toget).size(); i++) {
				if (parent[ad.get(toget).get(i)] == -1) {
					parent[ad.get(toget).get(i)] = toget;
					queue.addLast(ad.get(toget).get(i));
				}
			}
			lv.add(toget);
		}
 
		child[0]++;
	}
 
	static void sort(int[] A) {
		int n = A.length;
		Random rnd = new Random();
		for (int i = 0; i < n; ++i) {
			int tmp = A[i];
			int randomPos = i + rnd.nextInt(n - i);
			A[i] = A[randomPos];
			A[randomPos] = tmp;
		}
		Arrays.sort(A);
	}
 
	static void sort(long[] A) {
		int n = A.length;
		Random rnd = new Random();
		for (int i = 0; i < n; ++i) {
			long tmp = A[i];
			int randomPos = i + rnd.nextInt(n - i);
			A[i] = A[randomPos];
			A[randomPos] = tmp;
		}
		Arrays.sort(A);
	}
 
	static String sort(String s) {
		Character ch[] = new Character[s.length()];
		for (int i = 0; i < s.length(); i++) {
			ch[i] = s.charAt(i);
		}
		Arrays.sort(ch);
		StringBuffer st = new StringBuffer("");
		for (int i = 0; i < s.length(); i++) {
			st.append(ch[i]);
		}
		return st.toString();
	}
 
	public static long gcd(long a, long b) {
		if (a == 0) {
			return b;
		}
		return gcd1(b % a, a);
	}
 
	static class FastReader {
		BufferedReader br;
		StringTokenizer st;
 
		public FastReader() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}
 
		String next() {
			while (st == null || !st.hasMoreElements()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}
 
		int nextInt() {
			return Integer.parseInt(next());
		}
 
		long nextLong() {
			return Long.parseLong(next());
		}
 
		double nextDouble() {
			return Double.parseDouble(next());
		}
 
		String nextLine() {
			String str = "";
			try {
				str = br.readLine();
			} catch (IOException e) {
				e.printStackTrace();
			}
			return str;
		}
	}
}