if we know the value of $$$\min(a,b)$$$, and query both $$$\min(a,c)$$$ and $$$\min(b,c)$$$, we can easily determine which value is the smallest among $$$a, b, c$$$ (let's assume it's $$$b$$$) and we still know $$$\min(a,c)$$$, so we can restart the process with the next value $$$d$$$ for a total of $$$2n - 1$$$ queries.

#include "bits/stdc++.h"
using namespace std;
 
int query(int i, int j) {
    cout << "? " << i << ' ' << j << "\n";
    cout.flush();
    int res;
    cin >> res;
    return res;
}
 
int main(){
    int n;
    cin >> n;
    int ab = query(1, 2), a = 1, b = 2;
    vector<int> sol(n + 1, -1);
    for(int c = 3; c <= n; ++c) {
        int ac = query(a, c), bc = query(b, c);
        if(ac < ab) {
            sol[c] = ac;
        }
        else if(ac > ab) {
            sol[b] = ab;
            b = c;
            ab = ac;
        }
        else {
            sol[a] = ac;
            a = c;
            ab = bc;
        }
    }
    sol[a] = sol[b] = ab;
    cout << "! ";
    for(int i = 1; i <= n; ++i) {
        cout << sol[i] << ' ';
    }
    cout.flush();
}

#include "bits/stdc++.h"
using namespace std;
 
int query(int i, int j) {
    cout << "? " << i << ' ' << j << "\n";
    cout.flush();
    int res;
    cin >> res;
    return res;
}
 
mt19937 rng(234981239);
 
signed main(){
    int n;
    cin >> n;
    vector<int> sol(n + 1, -1), people(n);
    iota(people.begin(), people.end(), 1);
    shuffle(people.begin(), people.end(), rng);
    int ab = query(people[0], people[1]), a = people[0], b = people[1];
    for(int i = 2; i < n; ++i) {
        int c = people[i];
        int ac = query(a, c);
        if(ac < ab) {
            sol[c] = ac;
        }
        else if(ac > ab) {
            sol[b] = ab;
            b = c;
            ab = ac;
        }
        else {
            sol[a] = ac;
            a = c;
            int bc = query(b, c);
            ab = bc;
        }
    }
    sol[a] = sol[b] = ab;
    cout << "! ";
    for(int i = 1; i <= n; ++i) {
        cout << sol[i] << ' ';
    }
    cout.flush();
}