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 CPP):

Solution:
#include <bits/stdc++.h>

using namespace std;


#define     mod             (int)1e9+7
#define     MOD             998244353
#define     ll              long long
#define     int              long long
#define     mp              make_pair
#define     deb(x)          cout<< x << endl
#define     inf        1e18
#define     F               first
#define     S               second
#define     pb              push_back
#define     pf              emplace_front
#define     all(p)          p.begin(), p.end()
#define     rep(i,a,b)      for( i=a;i<b;++i)
#define     repr(i,a,b)     for( i=a;i>=b;i--)
#define     up              upper_bound
#define     lb              lower_bound
#define     t()             int test;cin>>test;while(test--)
#define     setbits(x)      __builtin_popcountll(x)
#define     si              set<int>
#define     que_max         priority_queue <int>
#define     que_min         priority_queue <int, vi, greater<int> >
#define     IOS             ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define     endl            "\n"
#define isOn(S, j)                   (S & (1 << j))
#define setBit(S, j)                 (S |= (1 << j))
#define clearBit(S, j)               (S &= ~(1 << j))
#define toggleBit(S, j)              (S ^= (1 << j))
#define lowBit(S)                    (S & (-S))
#define setAll(S, n)                 (S = (1 << n) - 1)
#define modulo(S, N)                 ((S) & (N - 1))
#define isPowerOfTwo(S)              (!(S & (S - 1)))
#define nearestPowerOfTwo(S)         ((int)pow(2.0, (int)((log((double)S) / log(2.0)) + 0.5)))
#define turnOffLastBit(S)            ((S) & (S - 1))
#define turnOnLastZero(S)            ((S) | (S + 1))
#define turnOffLastConsecutiveBits(S) ((S) & (S + 1))
#define turnOnLastConsecutiveZeroes(S) ((S) | (S - 1))

typedef pair<int, int>  pii;
typedef pair<ll, ll>  pl;
typedef vector<int>   vi;
typedef vector<ll>    vl;
typedef vector<pii>   vpii;
typedef vector<pl>    vpl;
typedef vector<vi>    vvi;
typedef vector<vl>    vvl;


#define TRACE
#ifdef TRACE
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cout << name << " : " << arg1 << endl;
  //use cerr if u want to display at the bottom
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ','); cout.write(names, comma - names) << " : " << arg1 << " | "; __f(comma + 1, args...);
}
#else
#define trace(...)
#endif
int mpow(int base, int exp);  
void ipgraph(int n, int m);
void dfs(int u, int par);

const int N = 100005, M = 100005;
//=======================

vl g[N];
int a[N];
ll n;
vl interact(ll r)
{
  cout<<"? "<<r<<endl;
  vl d(n+1);
  ll i;
  rep(i,1,n+1)
  cin>>d[i];
  return d;
}
void solve()
{
    ll i,j,m,k;
    cin>>n;
    if(n==1)
    {
      cout<<"! "<<endl;
      return;
    }
    int node=1;
    vl vis(n+1);
    set<pl> edge;
    vl arr=interact(node);
    ll odd=0,even=0;
    rep(i,1,n+1)
    {
      if(arr[i]==1)
        edge.insert({1,i});
      if(arr[i]%2)
        odd++;
      else
        even++;
    }
    // trace(odd,even);
    if(odd<even)
    {
      rep(i,2,n+1)
      {
        if(arr[i]%2==0)
          continue;
        vl tmp=interact(i);
        rep(j,1,n+1)
        {
          if(tmp[j]==1&&edge.find({j,i})==edge.end())
            edge.insert({i,j});
        }
      }
    }
    else
    {
       rep(i,2,n+1)
      {
        if(arr[i]%2)
          continue;
        vl tmp=interact(i);
        rep(j,1,n+1)
        {
          if(tmp[j]==1&&edge.find({j,i})==edge.end())
            edge.insert({i,j});
        }
      }
    }
    cout<<"! "<<endl;
    for(auto it:edge)
      cout<<it.F<<" "<<it.S<<endl;
}

int32_t main()
{
  // IOS;

  // t()
  {
       solve();
  }

}
int mpow(int base, int exp) {
  base %= (mod);
  int result = 1;
  while (exp > 0) {
    if (exp & 1) result = ((ll)result * base) % (mod);
    base = ((ll)base * base) % (mod);
    exp >>= 1;
  }
  return result;
}

void ipgraph(int n, int m){
  int i, u, v;
  while(m--){
    cin>>u>>v;
    u--, v--;
    g[u].pb(v);
    g[v].pb(u);
  }
}

void dfs(int u, int par){
  for(int v:g[u]){
    if (v == par) continue;
    dfs(v, u);
  }
}