description: 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.
solution: import java.util.*;
  import java.io.*;
  import java.math.*;
  
  public class cf {
  	static PrintWriter pw = new PrintWriter(System.out);
  
  	public static void main(String[] args) throws IOException, InterruptedException {
  		Scanner sc = new Scanner(System.in);
  		int n = sc.nextInt();
  		int[] arr = new int[n];
  		int c = 2;
  		while (c <= n) {
  			pw.print("? ");
  			for (int i = 0; i < n; i++) {
  				pw.print((i == n - 1 ? 1 : c) + " ");
  			}
  			pw.println();
  			pw.flush();
  			int x = sc.nextInt();
  			if (x == 0) {
  				break;
  			}
  			arr[x - 1] = -c + 1;
  			c++;
  		}
  		c = n - 1;
  		while (c > 0) {
  			pw.print("? ");
  			for (int i = 0; i < n; i++) {
  				pw.print((i == n - 1 ? n : c) + " ");
  			}
  			pw.println();
  			pw.flush();
  			int x = sc.nextInt();
  			if (x == 0) {
  				break;
  			}
  			arr[x - 1] = n - c;
  			c--;
  		}
  		int min = 200;
  		for (int i = 0; i < arr.length; i++) {
  			min = Math.min(min, arr[i]);
  		}
  		for (int i = 0; i < arr.length; i++) {
  			arr[i] += -min + 1;
  		}
  		pw.print("! ");
  		for (int i = 0; i < arr.length; i++) {
  			pw.print(arr[i] + " ");
  		}
  		pw.close();
  	}
  
  	public static class tuble implements Comparable<tuble> {
  		int x;
  		int y;
  		int z;
  
  		public tuble(int x, int y, int z) {
  			this.x = x;
  			this.y = y;
  			this.z = z;
  		}
  
  		public String toString() {
  			return x + " " + y + " " + z;
  		}
  
  		public int compareTo(tuble other) {
  			if (this.x == other.x) {
  				if (this.y == other.y)
  					return this.z - other.z;
  				return this.y - other.y;
  			} else {
  				return this.x - other.x;
  			}
  		}
  	}
  
  	public static class pair implements Comparable<pair> {
  		int x;
  		int y;
  
  		public pair(int x, int y) {
  			this.x = x;
  			this.y = y;
  		}
  
  		public String toString() {
  			return x + " " + y;
  		}
  
  		public boolean equals(Object o) {
  			if (o instanceof pair) {
  				pair p = (pair) o;
  				return p.x == x && p.y == y;
  			}
  			return false;
  		}
  
  		public int hashCode() {
  			return new Integer(x).hashCode() * 31 + new Integer(y).hashCode();
  		}
  
  		public int compareTo(pair other) {
  			if (this.x == other.x) {
  				return Long.compare(this.y, other.y);
  			}
  			return Long.compare(this.x, other.x);
  		}
  	}
  
  	static class Scanner {
  		StringTokenizer st;
  		BufferedReader br;
  
  		public Scanner(InputStream s) {
  			br = new BufferedReader(new InputStreamReader(s));
  		}
  
  		public Scanner(FileReader r) {
  			br = new BufferedReader(r);
  		}
  
  		public String next() throws IOException {
  			while (st == null || !st.hasMoreTokens())
  				st = new StringTokenizer(br.readLine());
  			return st.nextToken();
  		}
  
  		public int nextInt() throws IOException {
  			return Integer.parseInt(next());
  		}
  
  		public long nextLong() throws IOException {
  			return Long.parseLong(next());
  		}
  
  		public String nextLine() throws IOException {
  			return br.readLine();
  		}
  
  		public double nextDouble() throws IOException {
  			return Double.parseDouble(next());
  		}
  
  		public boolean ready() throws IOException {
  			return br.ready();
  		}
  
  	}
  
  }