Problem: This is an interactive problem!

Ehab plays a game with Laggy. Ehab has 2 hidden integers (a,b). Laggy can ask a pair of integers (c,d) and Ehab will reply with:

  * 1 if a ⊕ c>b ⊕ d. 
  * 0 if a ⊕ c=b ⊕ d. 
  * -1 if a ⊕ c<b ⊕ d. 



Operation a ⊕ b is the [bitwise-xor operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of two numbers a and b.

Laggy should guess (a,b) with at most 62 questions. You'll play this game. You're Laggy and the interactor is Ehab.

It's guaranteed that 0 ≤ a,b<2^{30}.

Input

See the interaction section.

Output

To print the answer, print "! a b" (without quotes). Don't forget to flush the output after printing the answer.

Interaction

To ask a question, print "? c d" (without quotes). Both c and d must be non-negative integers less than 2^{30}. Don't forget to flush the output after printing any question.

After each question, you should read the answer as mentioned in the legend. If the interactor replies with -2, that means you asked more than 62 queries and your program should terminate.

To flush the output, you can use:-

  * fflush(stdout) in C++. 
  * System.out.flush() in Java. 
  * stdout.flush() in Python. 
  * flush(output) in Pascal. 
  * See the documentation for other languages. 



Hacking:

To hack someone, print the 2 space-separated integers a and b (0 ≤ a,b<2^{30}).

Example

Input

1
-1
0

Output

? 2 1
? 1 2
? 2 0
! 3 1

Note

In the sample:

The hidden numbers are a=3 and b=1.

In the first query: 3 ⊕ 2 = 1 and 1 ⊕ 1 = 0, so the answer is 1.

In the second query: 3 ⊕ 1 = 2 and 1 ⊕ 2 = 3, so the answer is -1.

In the third query: 3 ⊕ 2 = 1 and 1 ⊕ 0 = 1, so the answer is 0.

Then, we printed the answer.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
bool DEBUGGING = 0;
void useiostream() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cerr.tie(0);
}
ll gcd(ll a, ll b) { return b == 0 ? a : gcd(b, a % b); }
ll lcm(ll a, ll b) { return a / gcd(a, b) * b; }
ld const pi = acos(-1.0);
ld const eps = 0.000000001;
template <class T>
bool mi(T &a, T b) {
  return a > b ? (a = b, true) : false;
}
template <class T>
bool ma(T &a, T b) {
  return a < b ? (a = b, true) : false;
}
void ans(int a, int b) {
  printf("! %d %d\n", a, b);
  fflush(stdout);
  exit(0);
}
int ask(int c, int d) {
  printf("? %d %d\n", c, d);
  fflush(stdout);
  int a;
  scanf("%d", &a);
  return a;
}
int main() {
  int pp = ask(0, 0);
  int a, b;
  a = b = 0;
  for (int j = 29; j >= 0; j--) {
    int p, q;
    p = ask((1 << j) ^ a, b);
    q = ask(a, (1 << j) ^ b);
    if (p == q) {
      if (pp == 1)
        a ^= 1 << j;
      else
        b ^= 1 << j;
      pp = p;
    } else {
      if (p == -1) a ^= 1 << j, b ^= 1 << j;
    }
  }
  ans(a, b);
  return 0;
}