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.io.*;
import java.util.*;
import java.math.BigInteger;

/**
 __                  __
 ( _)                ( _)
 / / \\              / /\_\_
 / /   \\            / / | \ \
 / /     \\          / /  |\ \ \
 /  /   ,  \ ,       / /   /|  \ \
 /  /    |\_ /|      / /   / \   \_\
 /  /  |\/ _ '_| \   / /   /   \    \\
 |  /   |/  0 \0\    / |    |    \    \\
 |    |\|      \_\_ /  /    |     \    \\
 |  | |/    \.\ o\o)  /      \     |    \\
 \    |     /\\`v-v  /        |    |     \\
 | \/    /_| \\_|  /         |    | \    \\
 | |    /__/_ `-` /   _____  |    |  \    \\
 \|    [__]  \_/  |_________  \   |   \    ()
 /    [___] (    \         \  |\ |   |   //
 |    [___]                  |\| \|   /  |/
 /|    [____]                  \  |/\ / / ||
 (  \   [____ /     ) _\      \  \    \| | ||
 \  \  [_____|    / /     __/    \   / / //
 |   \ [_____/   / /        \    |   \/ //
 |   /  '----|   /=\____   _/    |   / //
 __ /  /        |  /   ___/  _/\    \  | ||
 (/-(/-\)       /   \  (/\/\)/  |    /  | /
 (/\/\)           /   /   //
 _________/   /    /
 \____________/    (


 @author NTUDragons-Reborn
 */

public class Solution{
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        Task solver = new Task();
        solver.solve(in, out);
        out.close();
    }
    // main solver
    static class Task{

        double eps= 0.00000001;
        static final int MAXN = 100005;
        static final int MOD= 1000000007;

        // stores smallest prime factor for every number
        static int spf[] = new int[MAXN];

        static boolean[] prime;

        Map<Integer,Set<Integer>> dp= new HashMap<>();

        // Calculating SPF (Smallest Prime Factor) for every
        // number till MAXN.
        // Time Complexity : O(nloglogn)
        public void sieve()
        {
            spf[1] = 1;
            for (int i=2; i<MAXN; i++)

                // marking smallest prime factor for every
                // number to be itself.
                spf[i] = i;

            // separately marking spf for every even
            // number as 2
            for (int i=4; i<MAXN; i+=2)
                spf[i] = 2;

            for (int i=3; i*i<MAXN; i++)
            {
                // checking if i is prime
                if (spf[i] == i)
                {
                    // marking SPF for all numbers divisible by i
                    for (int j=i*i; j<MAXN; j+=i)

                        // marking spf[j] if it is not
                        // previously marked
                        if (spf[j]==j)
                            spf[j] = i;
                }
            }
        }

        void sieveOfEratosthenes(int n)
        {
            // Create a boolean array
            // "prime[0..n]" and
            // initialize all entries
            // it as true. A value in
            // prime[i] will finally be
            // false if i is Not a
            // prime, else true.
            prime= new boolean[n + 1];
            for (int i = 0; i <= n; i++)
                prime[i] = true;
    
            for (int p = 2; p * p <= n; p++)
            {
                // If prime[p] is not changed, then it is a
                // prime
                if (prime[p] == true)
                {
                    // Update all multiples of p
                    for (int i = p * p; i <= n; i += p)
                        prime[i] = false;
                }
            }
        }

        // A O(log n) function returning primefactorization
        // by dividing by smallest prime factor at every step
        public Set<Integer> getFactorization(int x)
        {
            if(dp.containsKey(x)) return dp.get(x);
            Set<Integer> ret = new HashSet<>();
            while (x != 1)
            {
                if(spf[x]!=2) ret.add(spf[x]);
                x = x / spf[x];
            }
            dp.put(x,ret);
            return ret;
        }
        // function to find first index >= x
        public int lowerIndex(List<Integer> arr, int n, int x)
        {
            int l = 0, h = n - 1;
            while (l <= h)
            {
                int mid = (l + h) / 2;
                if (arr.get(mid) >= x)
                    h = mid - 1;
                else
                    l = mid + 1;
            }
            return l;
        }
        public int lowerIndex(int[] arr, int n, int x)
        {
            int l = 0, h = n - 1;
            while (l <= h)
            {
                int mid = (l + h) / 2;
                if (arr[mid] >= x)
                    h = mid - 1;
                else
                    l = mid + 1;
            }
            return l;
        }

        // function to find last index <= y
        public int upperIndex(List<Integer> arr, int n, int y)
        {
            int l = 0, h = n - 1;
            while (l <= h)
            {
                int mid = (l + h) / 2;
                if (arr.get(mid) <= y)
                    l = mid + 1;
                else
                    h = mid - 1;
            }
            return h;
        }
        public int upperIndex(int[] arr, int n, int y)
        {
            int l = 0, h = n - 1;
            while (l <= h)
            {
                int mid = (l + h) / 2;
                if (arr[mid] <= y)
                    l = mid + 1;
                else
                    h = mid - 1;
            }
            return h;
        }

        // function to count elements within given range
        public int countInRange(List<Integer> arr, int n, int x, int y)
        {
            // initialize result
            int count = 0;
            count = upperIndex(arr, n, y) -
                    lowerIndex(arr, n, x) + 1;
            return count;
        }

        public int _gcd(int a,  int  b)
        {

            if(b == 0) {
                return a;
            }
            else {
                return _gcd(b, a % b);
            }
        }

        public int add(int a, int b){
            a+=b;
            if(a>=MOD) a-=MOD;
            else if(a<0) a+=MOD;
            return a;
        }

        public int mul(int a, int b){
            long res= (long)a*(long)b;
            return (int)(res%MOD);
        }

        public int power(int a, int b) {
            int ans=1;
            while(b>0){
                if((b&1)!=0) ans= mul(ans,a);
                b>>=1;
                a= mul(a,a);
            }
            return ans;
        }

        int[] fact= new int[MAXN];
        int[] inv= new int[MAXN];

        public int Ckn(int n, int k){
            if(k<0 || n<0) return 0;
            return mul(mul(fact[n],inv[k]),inv[n-k]);
        }

        public int inverse(int a){
            return power(a,MOD-2);
        }

        public void preprocess() {
            fact[0]=1;
            for(int i=1;i<MAXN;i++) fact[i]= mul(fact[i-1],i);
            inv[MAXN-1]= inverse(fact[MAXN-1]);
            for(int i=MAXN-2;i>=0;i--){
                inv[i]= mul(inv[i+1],i+1);
            }
        }

        /**
         * return VALUE of lower bound for unsorted array
         */
        public int lowerBoundNormalArray(int[] arr, int x){
            TreeSet<Integer> set= new TreeSet<>();
            for(int num: arr) set.add(num);
            return set.lower(x);
        }
        /**
         * return VALUE of upper bound for unsorted array
         */
        public int upperBoundNormalArray(int[] arr, int x){
            TreeSet<Integer> set= new TreeSet<>();
            for(int num: arr) set.add(num);
            return set.higher(x);
        }

        public void debugArr(int[] arr){
            for(int i: arr) out.print(i+" ");
            out.println();
        }
        
        InputReader in; PrintWriter out;
        Scanner sc= new Scanner(System.in);
        public void solve(InputReader in, PrintWriter out) {
            this.in=in; this.out=out;
            int t=1;
            while(t-->0){
                solveD();
            }
        }

        List<Pair> adj= new ArrayList<>();
        int n;
        private int[] ask(int x){
            System.out.println("? "+x);
            System.out.flush();
            int[] res= new int[n+1];
            for(int i=1;i<=n;i++) res[i]= sc.nextInt();
            return res;
        }
        public void output(){
            System.out.println("!");
            for(Pair p: adj) System.out.println(p.x+" "+p.y);
            System.out.flush();
        }

        public void solveD(){
            n= in.nextInt();
            int[] d= ask(1);
            List<Integer> odd= new ArrayList<>(), even= new ArrayList<>();
            for(int i=2;i<=n;i++) {
                if (d[i]%2==0) even.add(i); else odd.add(i);
                if(d[i]==1) adj.add(new Pair(1,i));
            }
            if(even.size()>odd.size()) {
                List<Integer> temp= even;
                even=odd;
                odd= temp;
            } 
            for(int u: even){
                int[] du= ask(u);
                for(int v=2;v<=n;v++) if(du[v]==1) adj.add(new Pair(u,v));
            }
            output();
        }

        
        

    }

    static class Venice{
        public Map<Long,Long> m= new HashMap<>();
        public long base=0;
        public long totalValue=0;
        private int M= 1000000007;

        private long addMod(long a, long b){
            a+=b;
            if(a>=M) a-=M;
            return a;
        }

        public void reset(){
            m= new HashMap<>();
            base=0;
            totalValue=0;
        }
        public void update(long add){
            base= base+ add;
        }
        public void add(long key, long val){
            long newKey= key-base;
            m.put(newKey, addMod(m.getOrDefault(newKey,(long)0),val));
        }
    }


    static class Tuple implements Comparable<Tuple>{
        int x, y, z;
        public Tuple(int x, int y, int z){
            this.x= x;
            this.y= y;
            this.z=z;
        }
        @Override
        public int compareTo(Tuple o){
            return this.x-o.x;
        }
    }

    static class Pair implements Comparable<Pair>{
        public int x;
        public int y;
        public Pair(int x, int y){
            this.x= x;
            this.y= y;
        }

        @Override
        public int compareTo(Pair o) {
            return this.x-o.x;
        }
    }

    // public static class compareL implements Comparator<Tuple>{
    //     @Override
    //     public int compare(Tuple t1, Tuple t2) {
    //         return t2.l - t1.l;
    //     }
    // }

    // fast input reader class;
    static class InputReader {
        BufferedReader br;
        StringTokenizer st;

        public InputReader(InputStream stream) {
            br = new BufferedReader(new InputStreamReader(stream));
        }

        public String nextToken() {
            while (st == null || !st.hasMoreTokens()) {
                String line = null;
                try {
                    line = br.readLine();
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
                if (line == null) {
                    return null;
                }
                st = new StringTokenizer(line);
            }
            return st.nextToken();
        }

        public int nextInt() {
            return Integer.parseInt(nextToken());
        }
        public double nextDouble(){
            return Double.parseDouble(nextToken());
        }
        public long nextLong(){
            return Long.parseLong(nextToken());
        }
        public int[] nextIntArr(int n){
            int[] arr= new int[n];
            for(int i=0;i<n;i++) arr[i]= nextInt();
            return arr;
        }
        public long[] nextLongArr(int n){
            long[] arr= new long[n];
            for(int i=0;i<n;i++) arr[i]= nextLong();
            return arr;
        }
        public List<Integer> nextIntList(int n){
            List<Integer> arr= new ArrayList<>();
            for(int i=0;i<n;i++) arr.add(nextInt());
            return arr;
        }
        public int[][] nextIntMatArr(int n, int m){
            int[][] mat= new int[n][m];
            for(int i=0;i<n;i++) for(int j=0;j<m;j++) mat[i][j]= nextInt();
            return mat;
        }
        public List<List<Integer>> nextIntMatList(int n, int m){
            List<List<Integer>> mat= new ArrayList<>();
            for(int i=0;i<n;i++){
                List<Integer> temp= new ArrayList<>();
                for(int j=0;j<m;j++) temp.add(nextInt());
                mat.add(temp);
            }
            return mat;
        }
        public char[] nextStringCharArr(){
            return nextToken().toCharArray();
        }
    }
}