Problem: Jury picked a polynomial f(x) = a_0 + a_1 ⋅ x + a_2 ⋅ x^2 + ... + a_k ⋅ x^k. k ≤ 10 and all a_i are integer numbers and 0 ≤ a_i < 10^6 + 3. It's guaranteed that there is at least one i such that a_i > 0.

Now jury wants you to find such an integer x_0 that f(x_0) ≡ 0 mod (10^6 + 3) or report that there is not such x_0.

You can ask no more than 50 queries: you ask value x_q and jury tells you value f(x_q) mod (10^6 + 3).

Note that printing the answer doesn't count as a query.

Interaction

To ask a question, print "? x_q" (0 ≤ x_q < 10^6 + 3). The judge will respond with a single integer f(x_q) mod (10^6 + 3). If you ever get a result of −1 (because you printed an invalid query), exit immediately to avoid getting other verdicts.

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. 



When you are ready to answer, print "! x_0" where x_0 is the answer or -1 if there is no such x_0.

You can ask at most 50 questions per test case.

Hack Format

To hack, use the following format.

The only line should contain 11 integers a_0, a_1, ..., a_{10} (0 ≤ a_i < 10^6 + 3, max_{0 ≤ i ≤ 10}{a_i} > 0) — corresponding coefficients of the polynomial.

Examples

Input


 
1000002

0


Output


? 0

? 1

! 1

Input


 
5

2

1



Output


? 2

? 1

? 0

! -1

Note

The polynomial in the first sample is 1000002 + x^2.

The polynomial in the second sample is 1 + x^2.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int DEBUG = 0;
const int MOD = 1000 * 1000 + 3;
mt19937 gen(
    (unsigned)chrono::high_resolution_clock::now().time_since_epoch().count());
#pragma comment(linker, "/STACK:67108864")
int bin_pow(int a, int pow) {
  int res = 1;
  while (pow) {
    if (pow & 1) res = (1LL * res * a) % MOD;
    a = (1LL * a * a) % MOD;
    pow /= 2;
  }
  return res;
}
int calc_inv(int x) { return bin_pow(x, MOD - 2); }
vector<int> gauss(vector<vector<int>> x, vector<int> y) {
  int n = y.size();
  for (int i = 0; i < n; i++) {
    int inv = calc_inv(x[i][i]);
    for (int j = i; j < n; j++) x[i][j] = (1LL * x[i][j] * inv) % MOD;
    y[i] = (1LL * y[i] * inv) % MOD;
    for (int j = i + 1; j < n; j++) {
      int mult = x[j][i];
      for (int k = 0; k < n; k++)
        x[j][k] = (x[j][k] - (1LL * x[i][k] * mult) % MOD + MOD) % MOD;
      y[j] = (y[j] - (1LL * y[i] * mult) % MOD + MOD) % MOD;
    }
  }
  vector<int> ans(n);
  for (int i = n - 1; i >= 0; i--) {
    ans[i] = y[i];
    for (int j = i + 1; j < n; j++)
      ans[i] = (ans[i] - (1LL * ans[j] * x[i][j]) % MOD + MOD) % MOD;
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  vector<vector<int>> x(11, vector<int>(11));
  vector<int> y(11);
  for (int i = 0; i < 11; i++) {
    int t = 1;
    for (int j = 0; j < 11; j++, t = (1LL * t * i) % MOD) x[i][j] = t;
    cout << "? " << i << endl;
    cin >> y[i];
  }
  vector<int> k = gauss(x, y);
  for (int i = 0; i < MOD; i++) {
    int val = 0;
    for (int j = 0, p = 1; j < k.size(); j++, p = (1LL * p * i) % MOD)
      val = (val + (1LL * p * k[j]) % MOD + MOD) % MOD;
    if (val == 0) {
      cout << "! " << i << endl;
      return 0;
    }
  }
  cout << "! -1" << endl;
  return 0;
}