Problem: This is an interactive problem.

Little Dormi was faced with an awkward problem at the carnival: he has to guess the edges of an unweighted tree of n nodes! The nodes of the tree are numbered from 1 to n.

The game master only allows him to ask one type of question:

  * Little Dormi picks a node r (1 ≤ r ≤ n), and the game master will reply with an array d_1, d_2, …, d_n, where d_i is the length of the shortest path from node r to i, for all 1 ≤ i ≤ n.



Additionally, to make the game unfair challenge Little Dormi the game master will allow at most ⌈n/2⌉ questions, where ⌈ x ⌉ denotes the smallest integer greater than or equal to x.

Faced with the stomach-churning possibility of not being able to guess the tree, Little Dormi needs your help to devise a winning strategy!

Note that the game master creates the tree before the game starts, and does not change it during the game.

Input

The first line of input contains the integer n (2 ≤ n ≤ 2 000), the number of nodes in the tree.

You will then begin interaction.

Output

When your program has found the tree, first output a line consisting of a single "!" followed by n-1 lines each with two space separated integers a and b, denoting an edge connecting nodes a and b (1 ≤ a, b ≤ n). Once you are done, terminate your program normally immediately after flushing the output stream.

You may output the edges in any order and an edge (a,b) is considered the same as an edge (b,a). Answering is not considered as a query.

Interaction

After taking input, you may make at most ⌈n/2⌉ queries. Each query is made in the format "? r", where r is an integer 1 ≤ r ≤ n that denotes the node you want to pick for that query.

You will then receive n space separated integers d_1, d_2, …, d_n, where d_i is the length of the shortest path from node r to i, followed by a newline.

After printing 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. 



If at any point you make an invalid query or try to make more than ⌈ n/2 ⌉ queries, the interaction will terminate immediately and you will receive a Wrong Answer verdict.

Hacks

To hack a solution, use the following format.

The first line contains the integer n (2 ≤ n ≤ 2 000).

The next n−1 lines contain two integers u and v (1 ≤ u,v ≤ n) denoting an edge between u and v (u ≠ v). These n-1 edges must form a tree.

Examples

Input


4

0 1 2 2

1 0 1 1

Output


? 1

? 2

!
4 2
1 2
2 3


Input


5

2 2 1 1 0


Output


? 5

!
4 5
3 5
2 4
1 3

Note

Here is the tree from the first example.

<image>

Notice that the edges can be output in any order.

Additionally, here are the answers for querying every single node in example 1:

  * 1: [0,1,2,2] 
  * 2: [1,0,1,1] 
  * 3: [2,1,0,2] 
  * 4: [2,1,2,0]



Below is the tree from the second example interaction.

<image>

Lastly, here are the answers for querying every single node in example 2:

  * 1: [0,4,1,3,2] 
  * 2: [4,0,3,1,2] 
  * 3: [1,3,0,2,1] 
  * 4: [3,1,2,0,1] 
  * 5: [2,2,1,1,0]

Solution (in JAVA):

Solution:
import java.util.*;
import java.io.*;
import java.math.BigInteger;
public class Codechef{
    //SOLUTION BEGIN
	
    void pre() throws Exception{}
    static long N=1000000007;
    class mycomp implements Comparator<pair>{
    	public int compare(pair a , pair b) {
    		if(a.x<b.x) {
    			return 1;
    		}
    		else if(a.x>b.x) {
    			return -1;
    		}
    		if(a.y<b.y) {
    			return 1;
    		}else if(a.y>b.y) {
    			return -1;
    		}
    		return 0;
    	}
    }
    void solve(int TC) throws Exception{
    	int n=ni();
    	System.out.println("? 1");
    	ArrayList<Integer> odd=new ArrayList<Integer>();
    	ArrayList<Integer> even=new ArrayList<Integer>();
    	int dis[]=new int[n+1];
    	ArrayList<pair> edges=new ArrayList<pair>();
    	dis[1]=ni();
    	for(int i=2;i<=n;i++) {
    		dis[i]=ni();
    		if(dis[i]%2==0) {
    			even.add(i);
    		}
    		else {
    			odd.add(i);
    		}
    		if(dis[i]==1) {
    			edges.add(new pair(1,i));
    		}
    		
    	}
    	if(even.size()> odd.size()) {
    		ArrayList<Integer> temp=even;
    		even=odd;
    		odd=temp;
    	}
    	int len=even.size();
    	for(int i=0;i<len;i++) {
    		int next=even.get(i);
    		System.out.println("? "+next);
    		for(int j=1;j<=n;j++) {
    			dis[j]=ni();
    			if(dis[j]==1) {
    				int a=next;
    				int b=j;
    				if(a>b) {
    					int temp=a;
    					a=b;
    					b=temp;
    				}
    				edges.add(new pair(a,b));
    			}
    		}
    	}
    	Collections.sort(edges,new mycomp());
    	System.out.println("!");
    	len = edges.size();
    	pair a=edges.get(0),b;
    	for(int i=1;i<len;i++) {
    		b=edges.get(i);
    		if(a.x==b.x && a.y==b.y) {
    			
    		}else {
    			pn(a.x+" "+a.y);
    		}
    		a=b;
    	}
    	pn(a.x+" "+a.y);
    	
    }
               
    //SOLUTION END
    void hold(boolean b)throws Exception{if(!b)throw new Exception("Hold right there, Sparky!");}
    static boolean multipleTC = false;
    FastReader in;PrintWriter out;
    void run() throws Exception{
        in = new FastReader();
        out = new PrintWriter(System.out);
        
        
    	
        int T = (multipleTC)?ni():1;
        pre();for(int t = 1; t<= T; t++)solve(t);
        out.flush();
        out.close();
    }
    public static void main(String[] args) throws Exception{
        new Codechef().run();
    }
    int bit(long n){return (n==0)?0:(1+bit(n&(n-1)));}
    void p(Object o){out.print(o);}
    void pn(Object o){out.println(o);}
    void pni(Object o){out.println(o);out.flush();}
    String n()throws Exception{return in.next();}
    String nln()throws Exception{return in.nextLine();}
    int ni()throws Exception{return Integer.parseInt(in.next());}
    long nl()throws Exception{return Long.parseLong(in.next());}
    double nd()throws Exception{return Double.parseDouble(in.next());}

    class FastReader{
        BufferedReader br;
        StringTokenizer st;
        public FastReader(){
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        public FastReader(String s) throws Exception{
            br = new BufferedReader(new FileReader(s));
        }

        String next() throws Exception{
            while (st == null || !st.hasMoreElements()){
                try{
                    st = new StringTokenizer(br.readLine());
                }catch (IOException  e){
                    throw new Exception(e.toString());
                }
            }
            return st.nextToken();
        }

        String nextLine() throws Exception{
            String str = "";
            try{   
                str = br.readLine();
            }catch (IOException e){
                throw new Exception(e.toString());
            }  
            return str;
        }
    }
    class pair{
    	int x, y;
    	pair(int x,int y){
    		this.x=x;
    		this.y=y;
    	}
    }
    
}