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 int N = 1010;
map<pair<int, long long>, long long> w;
int n;
long long L, each;
pair<long long, long long> ans[N];
long long ask(int id, long long x) {
  if (w[make_pair(id, x)] > 0) return w[make_pair(id, x)];
  cout << "? " << id << " " << x << endl;
  long long ret;
  cin >> ret;
  return w[make_pair(id, x)] = ret;
}
long long get(int id, long long req) {
  long long low = 0, high = 1e18, ans = high;
  while (low <= high) {
    long long mid = (low + high) >> 1;
    if (ask(id, mid) >= req) {
      ans = mid;
      high = mid - 1;
    } else {
      low = mid + 1;
    }
  }
  return ans;
}
int kth(vector<int> vec, long long req, int k) {
  if (vec.size() == 1) return vec.back();
  int p = (1LL * rand() * rand() + rand()) % vec.size();
  long long at = get(vec[p], req);
  assert(vec.size());
  vector<int> s, g, e(1, vec[p]);
  for (int id : vec)
    if (id != vec[p]) {
      long long val = ask(id, at);
      if (val < req) {
        g.push_back(id);
      } else if (val > req) {
        s.push_back(id);
      } else
        e.push_back(id);
    }
  if (s.size() + e.size() >= k && s.size() < k) return vec[p];
  if (k <= s.size()) return kth(s, req, k);
  return kth(g, req, k - s.size() - e.size());
}
void solve(vector<int> vec, int l, int r, long long X, long long Y) {
  if (vec.size() == 1) {
    ans[vec.back()].first = X;
    ans[vec.back()].second = Y;
    return;
  }
  int mid = vec.size() >> 1;
  long long val_mid = (l + mid - 1) * each;
  int p = kth(vec, val_mid, mid);
  long long at = get(p, val_mid);
  vector<int> s, g, e;
  for (int id : vec) {
    long long val = ask(id, at);
    if (val < val_mid) {
      g.push_back(id);
    } else if (val > val_mid)
      s.push_back(id);
    else {
      e.push_back(id);
    }
  }
  while (s.size() < mid) {
    s.push_back(e.back());
    e.pop_back();
  }
  while (e.size()) {
    g.push_back(e.back());
    e.pop_back();
  }
  solve(s, l, l + mid - 1, X, at);
  solve(g, l + mid, r, at, Y);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  srand(time(NULL));
  cin >> n >> L;
  each = L / n;
  vector<int> id;
  for (int i = 1; i <= n; ++i) id.push_back(i);
  solve(id, 1, n, 0, (long long)1e18);
  cout << "!" << endl;
  for (int i = 1; i <= n; ++i)
    cout << ans[i].first << " " << ans[i].second << endl;
  return 0;
}