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;
int n;
long long int L, al[1000], ar[1000];
map<long long int, long long int> M[1000], M2[1000];
long long int query(int i, long long int x) {
  if (M[i].count(x)) return M[i][x];
  long long int y;
  cout << "? " << i + 1 << " " << x << endl;
  cin >> y;
  M[i][x] = y, M2[i][y] = x;
  return y;
}
long long int pos(int i, long long int y) {
  auto it = M2[i].lower_bound(y);
  if ((it != M2[i].end()) && (it->first == y)) return it->second;
  long long int s = 0, e = 1e18;
  if (it != M2[i].end()) e = it->second - 1;
  if (it != M2[i].begin()) it--, s = it->second + 1;
  while (s < e) {
    long long int m = (s + e) / 2;
    if (query(i, m) < y)
      s = m + 1;
    else
      e = m;
  }
  return s;
}
int getMid(long long int m, int q, vector<int> v) {
  int t = 0;
  while (v.size() > 1) {
    t++;
    if (t > 20) break;
    random_shuffle(v.begin(), v.end());
    int i, p = 1;
    while ((1 << p) <= v.size()) p++;
    p /= 2, p = max(p, 2), p = min(p, (int)v.size());
    vector<pair<long long int, int> > vv;
    for (i = 0; i < p; i++) vv.push_back(make_pair(pos(v[i], m), v[i]));
    sort(vv.begin(), vv.end());
    vector<int> vvv[15];
    for (i = 0; i < vv.size(); i++) vvv[i].push_back(vv[i].second);
    for (i = p; i < v.size(); i++) {
      int ll = 0, rr = p;
      while (ll < rr) {
        int mm = (ll + rr) / 2;
        if (query(v[i], vv[mm].first) < m)
          ll = mm + 1;
        else
          rr = mm;
      }
      vvv[ll].push_back(v[i]);
    }
    int s = 0;
    for (i = 0; i < p + 1; i++) {
      s += vvv[i].size();
      if (s >= q) {
        s -= vvv[i].size();
        break;
      }
    }
    v = vvv[i], q -= s;
  }
  return v[0];
}
int findAns(long long int s, long long int e, long long int l, long long int r,
            vector<int> v) {
  if (v.size() == 1) {
    al[v[0]] = s, ar[v[0]] = e;
    return 0;
  }
  int i;
  vector<int> ll, mm, rr;
  long long int m = (l / (L / n) + r / (L / n)) / 2 * (L / n);
  int x = getMid(m, v.size() / 2, v);
  long long int y = pos(x, m);
  for (i = 0; i < v.size(); i++) {
    if (query(v[i], y) == m)
      mm.push_back(v[i]);
    else if (query(v[i], y) < m)
      rr.push_back(v[i]);
    else
      ll.push_back(v[i]);
  }
  while (!mm.empty()) {
    if (ll.size() >= rr.size())
      rr.push_back(mm.back());
    else
      ll.push_back(mm.back());
    mm.pop_back();
  }
  findAns(s, y, l, m, ll), findAns(y, e, m, r, rr);
  return 0;
}
int main() {
  int i;
  cin >> n >> L;
  vector<int> v;
  for (i = 0; i < n; i++) v.push_back(i);
  findAns(0, 1e18, 0, L, v);
  cout << "!" << endl;
  for (i = 0; i < n; i++) cout << al[i] << " " << ar[i] << endl;
  return 0;
}