#include <bits/stdc++.h>
using namespace std;

//高速化 
struct ponjuice{ponjuice(){cin.tie(0);ios::sync_with_stdio(0);cout<<fixed<<setprecision(20);}}PonJuice;
//#define endl '\n' //インタラクティブ問題の時は消す

//型
using ll = long long;
using ld = long double;
template<class T>using vc = vector<T>; template<class T>using vvc = vc<vc<T>>; template<class T>using vvvc = vvc<vc<T>>;
using vi = vc<int>;  using vvi = vvc<int>;  using vvvi = vvvc<int>;
using vl = vc<ll>;   using vvl = vvc<ll>;   using vvvl = vvvc<ll>;
using pi = pair<int, int>;  using pl = pair<ll, ll>;
using ull = unsigned ll;
template<class T>using priq = priority_queue<T>;
template<class T>using priqg = priority_queue<T, vc<T>, greater<T>>;

// for文
#define overload4(a, b, c, d, e, ...) e
#define rep1(n)             for(ll i = 0; i < n; i++)
#define rep2(i, n)          for(ll i = 0; i < n; i++)
#define rep3(i, a, b)       for(ll i = a; i < b; i++)
#define rep4(i, a, b, step) for(ll i = a; i < b; i+= step)
#define rep(...) overload4(__VA_ARGS__, rep4, rep3, rep2, rep1)(__VA_ARGS__)
#define per1(n)             for(ll i = n-1; i >= 0; i--)
#define per2(i, n)          for(ll i = n-1; i >= 0; i--)
#define per3(i, a, b)       for(ll i = b-1; i >= a; i--)
#define per4(i, a, b, step) for(ll i = b-1; i >= a; i-= step)
#define per(...) overload4(__VA_ARGS__, per4, per3, per2, per1)(__VA_ARGS__)

//関数
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define si(x) (ll)(x).size()
template<class S, class T>inline bool chmax(S& a, T b){return a < b && ( a = b , true);}
template<class S, class T>inline bool chmin(S& a, T b){return a > b && ( a = b , true);}

inline void yes(){cout << "Yes\n";}
inline void no(){cout << "No\n";}
inline void yesno(bool y = true){if(y)yes();else no();}

//定数
constexpr ll mod = 998244353;
constexpr ll minf=-(1<<29);
constexpr ll inf=(1<<29);
constexpr ll MINF=-(1LL<<60);
constexpr ll INF=(1LL<<60);
const int dx[4] ={-1, 0, 1, 0};
const int dy[4] ={ 0, 1, 0,-1};
const int dx8[8] ={-1,-1,-1, 0, 1, 1, 1, 0};
const int dy8[8] ={-1, 0, 1, 1, 1, 0,-1,-1};

void solve();
int main() {
	int t = 1;
    // cin >>t;
    while(t--)solve();
}

/*
せっかくなので、PPC は全てコメント書きます


インタラクティブ多めでいいね!

確実にゼロでは無い位があるので、それを用いて聞いていく

0でない返答が2個以上ある場合、
ab, ac, bc の情報から一位に答えが定まる
具体的には
sqrt(ab*bc*ca)/ab = c なので、後はノリで


0 である返答が1つしかない場合
81,64,47,36,25,1 の場合のみ一意に定まりそう


0この場合
不可能
*/

ll sqrt_ll(ll x) {
    ll res = sqrtl(x);
    while(res * res < x) res++;
    while(res * res > x) res--;
    return res;
}

void solve(){
    int n;
    cin >> n;
    vector<int> res(n-1, 0);
    int cnt = 0;
    rep(i,0,n-1) {
        cout << "? " << i << " " << n-1 << endl;
        cin >> res[i];
        if(res[i]) cnt++;
    }
    if(cnt == 0) {
        cout << "! -1" << endl;
        return;
    }
    if(cnt == 1) {
        set<int> ok = {81,64,47,36,25,1};
        rep(i,0,n-1) {
            if(ok.count(res[i])) {
                ll sq = sqrt_ll(res[i]);

                cout << "! ";
                rep(j,0,n) {
                    if(j == 0 || j == n-1-i) cout << sq;
                    else cout << 0;
                }
                cout << endl;

                return;
            }
        } 
        cout << "! -1" << endl;
        return;
    }

    int x = -1,y = 0;
    rep(i,0,n-1) {
        if(res[i]) {
            if(x == -1) x = i;
            else y = i;
        }
    }
    cout << "? " << x << " " << y << endl;
    int rs;
    cin >> rs;

    ll sq = rs * res[x] * res[y];
    sq = sqrt_ll(sq);
    ll a = sq / rs;
    
    cout << "! ";
    cout << a;
    rep(i,0,n-1) cout << res[n-2-i]/a;
    cout << endl;
}