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:
///*بِسْمِ اللَّهِ الرَّحْمَنِ الرَّحِيم*///

//#pragma GCC optimize("Ofast")
//#pragma GCC target("avx,avx2,fma")
//#pragma GCC optimization ("unroll-loops")

#include <bits/stdc++.h>
using namespace std;
#define ln '\n'
#define inp(x) scanf("%lld",&x)
#define inps(x) scanf("%s",x)
#define inp2(a,b) scanf("%lld %lld",&a,&b)
#define No              cout<<"No\n"
#define Yes             cout<<"Yes\n"
#define no              cout<<"no\n"
#define yes             cout<<"yes\n"
#define NO              cout<<"NO\n"
#define YES             cout<<"YES\n"

#define f0(i,b) for(int i=0;i<(b);i++)
#define f1(i,b) for(int i=1;i<=(b);i++)

#define MOD 1000000007
#define PI acos(-1)
#define ll long long int
typedef pair<int,int> pii;
typedef pair<ll, ll> pll;
#define pb push_back
#define F first
#define S second
#define sz size()
#define LCM(a,b) (a*(b/__gcd(a,b)))
//#define harmonic(n) 0.57721566490153286060651209+log(n)+1.0/(2*n)
#define SORT_UNIQUE(c) (sort(c.begin(),c.end()), c.resize(distance(c.begin(),unique(c.begin(),c.end()))))
#define all(v) v.begin(),v.end()
#define MEM(a, b) memset(a, b, sizeof(a))

#define fastio  ios_base::sync_with_stdio(false);
///Inline functions

inline bool EQ(double a, double b)
{
    return fabs(a-b) < 1e-9;
}
//inline bool isLeapYll year) { return (year%400==0) | (year%4==0 && year%100!=0); }
inline void normal(ll &a)
{
    a %= MOD;
    (a < 0) && (a += MOD);
}
inline ll modMul(ll a, ll b)
{
    a %= MOD, b %= MOD;
    normal(a), normal(b);
    return (a*b)%MOD;
}
inline ll modAdd(ll a, ll b)
{
    a %= MOD, b %= MOD;
    normal(a), normal(b);
    return (a+b)%MOD;
}
inline ll modSub(ll a, ll b)
{
    a %= MOD, b %= MOD;
    normal(a), normal(b);
    a -= b;
    normal(a);
    return a;
}
inline ll modPow(ll b, ll p)
{
    ll r = 1;
    while(p)
    {
        if(p&1)
            r = modMul(r, b);
        b = modMul(b, b);
        p >>= 1;
    }
    return r;
}
inline ll modInverse(ll a)
{
    return modPow(a, MOD-2);
}
inline ll modDiv(ll a, ll b)
{
    return modMul(a, modInverse(b));
}
inline bool isInside(pii p,ll n,ll m)
{
    return (p.first>=0&&p.first<n&&p.second>=0&&p.second<m);
}
inline bool isInside(pii p,ll n)
{
    return (p.first>=0&&p.first<n&&p.second>=0&&p.second<n);
}
inline bool isSquare(ll x)
{
    ll s = sqrt(x);
    return (s*s==x);
}
inline bool isFib(ll x)
{
    return isSquare(5*x*x+4)|| isSquare(5*x*x-4);
}
inline bool isPowerOfTwo(ll x)
{
    return ((1LL<<(ll)log2(x))==x);
}


/// DEBUG --------------------------------------------------------------------------------->>>>>>

///**
template < typename F, typename S >
ostream& operator << ( ostream& os, const pair< F, S > & p )
{
    return os << "(" << p.first << ", " << p.second << ")";
}

template < typename T >
ostream &operator << ( ostream & os, const vector< T > &v )
{
    os << "{";
    for (auto it = v.begin(); it != v.end(); ++it)
    {
        if ( it != v.begin() )
            os << ", ";
        os << *it;
    }
    return os << "}";
}

template < typename T >
ostream &operator << ( ostream & os, const set< T > &v )
{
    os << "[";
    for (auto it = v.begin(); it != v.end(); ++it)
    {
        if ( it != v.begin())
            os << ", ";
        os << *it;
    }
    return os << "]";
}

template < typename F, typename S >
ostream &operator << ( ostream & os, const map< F, S > &v )
{
    os << "[";
    for (auto it = v.begin(); it != v.end(); ++it)
    {
        if ( it != v.begin() )
            os << ", ";
        os << it -> first << " = " << it -> second ;
    }
    return os << "]";
}

#define dbg(args...) do {cerr << #args << " : "; faltu(args); } while(0)

clock_t tStart = clock();
#define timeStamp dbg("Execution Time: ", (double)(clock() - tStart)/CLOCKS_PER_SEC)

void faltu ()
{
    cerr << endl;
}

template <typename T>
void faltu( T a[], int n )
{
    for (int i = 0; i < n; ++i)
        cerr << a[i] << ' ';
    cerr << endl;
}

template <typename T, typename ... hello>
void faltu( T arg, const hello &... rest)
{
    cerr << arg << ' ';
    faltu(rest...);
}


/// TEMPLATE ----------------------------------------------------------------------->>>>>>

/*------------------------------Graph Moves----------------------------*/
//Rotation: S -> E -> N -> W
//const int fx[] = {0, +1, 0, -1};
//const int fy[] = {-1, 0, +1, 0};
//const int fx[]={+0,+0,+1,-1,-1,+1,-1,+1};   // Kings Move
//const int fy[]={-1,+1,+0,+0,+1,+1,-1,-1};  // Kings Move
//const int fx[]={-2, -2, -1, -1,  1,  1,  2,  2};  // Knights Move
//const int fy[]={-1,  1, -2,  2, -2,  2, -1,  1}; // Knights Move
/*---------------------------------------------------------------------*/

/// Bit Operations

/// inline bool checkBit(ll n, int i) { return n&(1LL<<i); }
/// inline ll setBit(ll n, int i) { return n|(1LL<<i); }
/// inline ll resetBit(ll n, int i) { return n&(~(1LL<<i)); }


/// ************************************** Code starts here ****************************************** */

/*
// Program showing a policy-based data structure.
#include <ext/pb_ds/assoc_container.hpp> // Common file
#include <ext/pb_ds/tree_policy.hpp>
#include <functional> // for less
using namespace __gnu_pbds;

// GNU link : https://goo.gl/WVDL6g
typedef tree<int, null_type, less_equal<int>, rb_tree_tag,
        tree_order_statistics_node_update>
        new_data_set;
// find_by_order(k) – ফাংশনটি kth ordered element এর একটা পয়েন্টার রিটার্ন করে। অর্থাৎ তুমি চাইলেই kth ইন্ডেক্সে কি আছে, সেটা জেনে ফেলতে পারছো!
// order_of_key(x) – ফাংশনটি x এলিমেন্টটা কোন পজিশনে আছে সেটা বলে দেয়।

void printOrderedSet(new_data_set s){
    //Function to show the contents of the set
    for(auto it=s.begin(); it!=s.end(); it++){
        cout<<*it<<" ";
    }cout<<endl;
}

void myerase(new_data_set &t, int v){
    int rank = t.order_of_key(v);//Number of elements that are less or equal to v in t
    new_data_set::iterator it = t.find_by_order(rank); //Iterator that points to the (rank)th element in t
    t.erase(it);
}

*/


/// randomely select pos in a range(l,r)
/*
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int pos = uniform_int_distribution<int>(l,r)(rng);



 freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
    freopen("error.txt", "w", stderr);

*/
const ll INF = 0x3f3f3f3f3f3f3f3f;
const long double EPS = 1e-9;
const int inf = 0x3f3f3f3f;
const int mx = (int)1e5+9;

ll n,m,a,b,t,i,j,d,cs=0,counT=0,k,ans=0,l=0,sum1=0,sum=0,Max,Min,num;

vector<int>Query(int i,int n)
{
    cout << "? " << i << ln;
    cout.flush();
    vector<int>ans(n+1);
    for(j=1; j<=n; j++)
        cin >> ans[j];
    return ans;
}

int main()
{
    fastio;

    cin >> n ;
    set<pii>vc;
    int odd = 0, even = 0 ;

    vector<int>ret = Query(1,n);
    for(j=2; j<=n; j++)
    {
        if(ret[j]==1)
            vc.insert({1,j});
        if(ret[j]&1)
            odd++;
        else
            even++;
    }

    if(odd < even)
    {
        for(i=2; i<=n; i++)
        {
            if(ret[i]&1)
            {
                vector<int>ret1 = Query(i,n);
                for(j=1; j<=n; j++)
                {
                    if(ret1[j]==1)
                        vc.insert({min(j,i),max(i,j)});
                }
            }
        }
    }
    else
    {
        for(i=2; i<=n; i++)
        {
            if(ret[i]%2==0)
            {
                vector<int>ret1 = Query(i,n);
                for(j=1; j<=n; j++)
                {
                    if(ret1[j]==1)
                        vc.insert({min(j,i),max(i,j)});
                }
            }
        }
    }

    cout<<"!"<<endl;
    for(auto x: vc)
      cout<<x.first<<" "<<x.second<<endl;
    cout.flush();
}