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 N = 1e3 + 2;
pair<long long, long long> ans[N];
long long n;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
long long ask(long long idx, long long val, long long debt) {
  cout << "? " << idx << ' ' << val << endl;
  cin >> idx;
  return idx - debt;
}
void solve(long long lef, long long rig, vector<long long> idx, long long val,
           long long debt) {
  if (!idx.size()) {
    return;
  }
  if (idx.size() == 1) {
    ans[idx[0]] = {lef, rig};
    return;
  }
  long long sz = idx.size() / 2, newval = (val / idx.size()) * sz, i, j, k, l,
            m, need = sz;
  vector<long long> temp, lis1, lis2, lis3, idx1, idx2;
  for (i = 0; i < idx.size(); i++) {
    temp.push_back(idx[i]);
  }
  while (temp.size() > 1) {
    i = temp[rng() % temp.size()];
    j = lef;
    k = rig;
    while (j < k) {
      m = (j + k) / 2;
      l = ask(i, m, debt);
      if (l >= newval) {
        k = m;
      } else {
        j = m + 1;
      }
    }
    lis1.clear();
    lis2.clear();
    lis3.clear();
    for (i = 0; i < temp.size(); i++) {
      j = ask(temp[i], k, debt);
      if (j > newval) {
        lis1.push_back(temp[i]);
      } else {
        if (j == newval) {
          lis3.push_back(temp[i]);
        } else {
          lis2.push_back(temp[i]);
        }
      }
    }
    if (lis1.size() <= need && lis2.size() <= temp.size() - need) {
      while (lis1.size() + 1 <= need) {
        lis1.push_back(lis3.back());
        lis3.pop_back();
      }
      while (lis3.size()) {
        lis2.push_back(lis3.back());
        lis3.pop_back();
      }
      while (lis1.size()) {
        idx1.push_back(lis1.back());
        lis1.pop_back();
      }
      while (lis2.size()) {
        idx2.push_back(lis2.back());
        lis2.pop_back();
      }
      break;
    } else {
      if (lis1.size() > need) {
        temp = lis1;
        while (lis2.size()) {
          idx2.push_back(lis2.back());
          lis2.pop_back();
        }
        while (lis3.size()) {
          idx2.push_back(lis3.back());
          lis3.pop_back();
        }
      } else {
        need -= (lis1.size() + lis3.size());
        temp = lis2;
        while (lis1.size()) {
          idx1.push_back(lis1.back());
          lis1.pop_back();
        }
        while (lis3.size()) {
          idx1.push_back(lis3.back());
          lis3.pop_back();
        }
      }
    }
  }
  solve(lef, k, idx1, newval, debt);
  solve(k, rig, idx2, (val / idx.size()) * (idx.size() - sz), debt + newval);
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long m, i, j, k, l;
  cin >> n >> l;
  vector<long long> idx;
  for (i = 1; i <= n; i++) {
    idx.push_back(i);
  }
  solve(0, 1e18, idx, l, 0);
  cout << "!" << endl;
  for (i = 1; i <= n; i++) {
    cout << ans[i].first << ' ' << ans[i].second << '\n';
  }
}