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;
ostream &operator<<(ostream &out, string str) {
  for (char c : str) out << c;
  return out;
}
template <class L, class R>
ostream &operator<<(ostream &out, pair<L, R> p) {
  return out << "(" << p.first << ", " << p.second << ")";
}
template <class T>
auto operator<<(ostream &out, T &&x) -> decltype(x.begin(), out) {
  out << '{';
  for (auto &e : x) out << e << (&e == &*--x.end() ? "" : ", ");
  return out << '}';
}
template <class... Args>
void dump(Args &&...args) {
  ((cerr << args << ";  "), ...);
}
template <class T>
int size(T &&a) {
  return (int)a.size();
}
using LL = long long;
using PII = pair<int, int>;
mt19937 rng(2137);
int rd(int a) { return rng() % a; }
int n;
LL L;
LL ask(int i, LL x) {
  cout << "? " << i + 1 << " " << x << endl;
  LL ret;
  cin >> ret;
  return ret;
}
LL search(int i, LL val, LL l, LL r) {
  while (l < r) {
    LL m = (l + r) / 2, a = ask(i, m);
    if (a == val)
      return m;
    else if (a < val)
      l = m + 1;
    else
      r = m - 1;
  }
  return l;
};
LL divide(vector<int> cur, LL val, LL l, LL r, vector<int> &low,
          vector<int> &high) {
  int goal_low = size(cur) / 2;
  int goal_high = size(cur) - goal_low;
  while (true) {
    int f = cur[rd(size(cur))];
    LL x = search(f, val, l, r);
    vector<int> left, mid, right;
    for (int i : cur) {
      LL a = ask(i, x);
      if (val < a)
        left.emplace_back(i);
      else if (val == a)
        mid.emplace_back(i);
      else
        right.emplace_back(i);
    }
    if (size(low) + size(left) <= goal_low) {
      for (int i : left) low.emplace_back(i);
      left.clear();
      while (size(mid) && size(low) < goal_low) {
        low.emplace_back(mid.back());
        mid.pop_back();
      }
    }
    if (size(right) + size(high) <= goal_high) {
      for (int i : right) high.emplace_back(i);
      right.clear();
      while (size(mid) && size(high) < goal_high) {
        high.emplace_back(mid.back());
        mid.pop_back();
      }
    }
    if (size(left))
      cur = left;
    else if (size(right))
      cur = right;
    else
      return x;
  }
}
vector<pair<LL, LL>> ans;
void solve(vector<int> cur, int done, LL l, LL r) {
  if (size(cur) == 1) {
    ans[cur.front()] = {l, r};
    return;
  }
  LL q = (L / n) * (done + size(cur) / 2);
  vector<int> low, high;
  LL p = divide(cur, q, l, r, low, high);
  solve(low, done, l, p);
  solve(high, done + size(low), p, r);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> L;
  ans.resize(n);
  vector<int> cur(n);
  for (int i = 0; i < n; i++) cur[i] = i;
  solve(cur, 0, 0, 1e18);
  cout << "!\n";
  for (auto &[l, r] : ans) cout << l << " " << r << "\n";
}