Problem: We call a function good if its domain of definition is some set of integers and if in case it's defined in x and x-1, f(x) = f(x-1) + 1 or f(x) = f(x-1).

Tanya has found n good functions f_{1}, …, f_{n}, which are defined on all integers from 0 to 10^{18} and f_i(0) = 0 and f_i(10^{18}) = L for all i from 1 to n. It's an notorious coincidence that n is a divisor of L. 

She suggests Alesya a game. Using one question Alesya can ask Tanya a value of any single function in any single point. To win Alesya must choose integers l_{i} and r_{i} (0 ≤ l_{i} ≤ r_{i} ≤ 10^{18}), such that f_{i}(r_{i}) - f_{i}(l_{i}) ≥ L/n (here f_i(x) means the value of i-th function at point x) for all i such that 1 ≤ i ≤ n so that for any pair of two functions their segments [l_i, r_i] don't intersect (but may have one common point).

Unfortunately, Tanya doesn't allow to make more than 2 ⋅ 10^{5} questions. Help Alesya to win!

It can be proved that it's always possible to choose [l_i, r_i] which satisfy the conditions described above.

It's guaranteed, that Tanya doesn't change functions during the game, i.e. interactor is not adaptive

Input

The first line contains two integers n and L (1 ≤ n ≤ 1000, 1 ≤ L ≤ 10^{18}, n is a divisor of L) — number of functions and their value in 10^{18}.

Output

When you've found needed l_i, r_i, print "!" without quotes on a separate line and then n lines, i-th from them should contain two integers l_i, r_i divided by space.

Interaction

To ask f_i(x), print symbol "?" without quotes and then two integers i and x (1 ≤ i ≤ n, 0 ≤ x ≤ 10^{18}). Note, you must flush your output to get a response.

After that, you should read an integer which is a value of i-th function in point x.

You're allowed not more than 2 ⋅ 10^5 questions.

To flush you can use (just after printing an integer and end-of-line):

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



Hacks:

Only tests where 1 ≤ L ≤ 2000 are allowed for hacks, for a hack set a test using following format:

The first line should contain two integers n and L (1 ≤ n ≤ 1000, 1 ≤ L ≤ 2000, n is a divisor of L) — number of functions and their value in 10^{18}.

Each of n following lines should contain L numbers l_1, l_2, ... , l_L (0 ≤ l_j < 10^{18} for all 1 ≤ j ≤ L and l_j < l_{j+1} for all 1 < j ≤ L), in i-th of them l_j means that f_i(l_j) < f_i(l_j + 1).

Example

Input


5 5
? 1 0
? 1 1
? 2 1
? 2 2
? 3 2
? 3 3
? 4 3
? 4 4
? 5 4
? 5 5
!
0 1
1 2
2 3
3 4
4 5


Output


0
1
1
2
2
3
3
4
4
4
5

Note

In the example Tanya has 5 same functions where f(0) = 0, f(1) = 1, f(2) = 2, f(3) = 3, f(4) = 4 and all remaining points have value 5.

Alesya must choose two integers for all functions so that difference of values of a function in its points is not less than L/n (what is 1 here) and length of intersection of segments is zero.

One possible way is to choose pairs [0, 1], [1, 2], [2, 3], [3, 4] and [4, 5] for functions 1, 2, 3, 4 and 5 respectively.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const long long NMAX = 1000;
pair<long long, long long> func[NMAX + 5];
pair<long long, long long> tmp_func[NMAX + 5];
long long initial[NMAX + 5];
pair<long long, long long> ans[NMAX + 5];
map<pair<long long, long long>, long long> memo;
long long singular_query(long long id, long long wh) {
  if (memo.count({id, wh}) == 0) {
    cout << "? " << id << " " << wh << endl;
    cin >> memo[{id, wh}];
  }
  return memo[{id, wh}];
}
void query_fun(long long fst, long long lst, long long wh) {
  for (long long i = fst; i <= lst; i++) {
    func[i].first = singular_query(func[i].second, wh);
  }
}
void solve(long long fst, long long lst, long long l, long long r,
           long long buck) {
  if (fst == lst) {
    ans[func[fst].second] = {l, r};
    return;
  }
  query_fun(fst, lst, l);
  for (long long i = fst; i <= lst; i++) {
    initial[func[i].second] = func[i].first;
  }
  long long target_split = (lst + fst) / 2;
  long long st = l + 1LL * (target_split - fst + 1) * buck - 1,
            dr = r - 1LL * (lst - target_split) * buck + 1;
  while (dr - st > 1) {
    long long mid = (st + dr) / 2;
    for (long long i = fst; i <= lst; i++) {
      map<pair<long long, long long>, long long>::iterator it =
          memo.lower_bound({func[i].second, mid});
      if (it != memo.end() && it->first.first == func[i].second &&
          it->second - initial[func[i].second] <
              buck * (target_split - fst + 1)) {
        func[i].first = it->second;
      } else if (it != memo.begin()) {
        it--;
        if (it->first.first == func[i].second &&
            it->second - initial[func[i].second] >=
                buck * (target_split - fst + 1)) {
          func[i].first = it->second;
        } else {
          func[i].first = singular_query(func[i].second, mid);
        }
      } else {
        func[i].first = singular_query(func[i].second, mid);
      }
    }
    sort(func + fst, func + 1 + lst,
         [&](pair<long long, long long> &a, pair<long long, long long> &b) {
           return a.first - initial[a.second] > b.first - initial[b.second];
         });
    if (func[target_split].first - initial[func[target_split].second] >=
        buck * (target_split - fst + 1)) {
      dr = mid;
      for (long long i = fst; i <= lst; i++) {
        tmp_func[i] = func[i];
      }
    } else {
      st = mid;
    }
  }
  for (long long i = fst; i <= lst; i++) {
    func[i] = tmp_func[i];
  }
  solve(fst, target_split, l, dr, buck);
  solve(target_split + 1, lst, dr, r, buck);
}
int main() {
  long long n;
  long long l;
  scanf("%lld %lld", &n, &l);
  for (long long i = 1; i <= n; i++) {
    func[i] = {0, i};
    memo[{i, 0}] = 0;
    memo[{i, 1e18}] = l;
  }
  solve(1, n, 0, 1e18, l / n);
  cout << "!" << endl;
  for (long long i = 1; i <= n; i++) {
    cout << ans[i].first << " " << ans[i].second << endl;
  }
  return 0;
}