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;
vector<pair<long long, long long> > ans;
long long dist;
long long inter(int i, long long x) {
  long long a = 0;
  cout << "? " << i << " " << x << endl;
  cin >> a;
  return a;
}
void divide(long long l, long long r, int mi, int ma, vector<int>& todo) {
  if (ma - mi == 1) {
    ans[todo[0]] = make_pair(l, r);
    return;
  }
  int next = (ma + mi) / 2;
  long long low = l + 1;
  long long high = r - 1;
  vector<int> big_todo, small_todo;
  vector<bool> poss(todo.size(), true);
  while (true) {
    int piv = rand() % todo.size();
    while (!poss[piv]) piv = rand() % todo.size();
    piv = todo[piv];
    long long low_piv = low;
    long long high_piv = high;
    long long tr;
    while (true) {
      tr = (low_piv + high_piv) / 2;
      long long ans_piv = inter(piv, tr);
      if (ans_piv == next * dist)
        break;
      else if (ans_piv < next * dist)
        low_piv = tr + 1;
      else
        high_piv = tr - 1;
    }
    int cnt_big = 0;
    int cnt_small = 0;
    vector<int> dir(todo.size());
    for (int i = 0; i < todo.size(); i++) {
      if (!poss[i]) continue;
      long long a = inter(todo[i], tr);
      if (a > next * dist) {
        cnt_big++;
        dir[i] = 1;
      } else if (a < next * dist) {
        cnt_small++;
        dir[i] = -1;
      }
    }
    if (cnt_big + big_todo.size() <= next - mi &&
        cnt_small + small_todo.size() <= ma - next) {
      for (int i = 0; i < todo.size(); i++) {
        if (!poss[i]) continue;
        if (dir[i] == 1) {
          big_todo.push_back(todo[i]);
          poss[i] = false;
        } else if (dir[i] == -1) {
          small_todo.push_back(todo[i]);
          poss[i] = false;
        }
      }
      for (int i = 0; i < todo.size(); i++) {
        if (!poss[i]) continue;
        if (big_todo.size() < next - mi) {
          big_todo.push_back(todo[i]);
          poss[i] = false;
        } else {
          small_todo.push_back(todo[i]);
          poss[i] = false;
        }
      }
      divide(l, tr, mi, next, big_todo);
      divide(tr, r, next, ma, small_todo);
      return;
    }
    if (cnt_big + big_todo.size() > next - mi) {
      high = tr - 1;
      for (int i = 0; i < todo.size(); i++) {
        if (poss[i] && dir[i] <= 0) {
          small_todo.push_back(todo[i]);
          poss[i] = false;
        }
      }
    } else if (cnt_small + small_todo.size() > ma - next) {
      low = tr + 1;
      for (int i = 0; i < todo.size(); i++) {
        if (poss[i] && dir[i] >= 0) {
          big_todo.push_back(todo[i]);
          poss[i] = false;
        }
      }
    }
  }
}
int main() {
  int n;
  long long L;
  cin >> n >> L;
  ans.resize(n + 1);
  dist = L / n;
  vector<int> todo(n);
  for (int i = 0; i < n; i++) todo[i] = i + 1;
  divide(0, 1000000000000000000, 0, n, todo);
  cout << "! " << endl;
  for (int i = 1; i <= n; i++) {
    cout << ans[i].first << " " << ans[i].second << endl;
  }
  return 0;
}