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;
using ll = long long;
using vi = vector<int>;
using vvi = vector<vi>;
using vll = vector<ll>;
using vvll = vector<vll>;
using vb = vector<bool>;
using vd = vector<double>;
using vs = vector<string>;
template <typename T>
void ckmin(T& a, const T& b) {
  a = min(a, b);
}
template <typename T>
void ckmax(T& a, const T& b) {
  a = max(a, b);
}
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
ll eval(int i, ll x) {
  cout << "? " << i + 1 << " " << x << endl;
  ll res;
  cin >> res;
  return res;
}
ll search(int i, ll xL, ll xH, ll v) {
  ll res = xH--;
  while (xL <= xH) {
    ll x = (xL + xH) / 2;
    if (eval(i, x) >= v)
      res = x, xH = x - 1;
    else
      xL = x + 1;
  }
  return res;
}
void solve(vector<array<ll, 2>>& ans, vi& elts, ll xL, ll xH, ll vL, ll vH) {
  if (int((elts).size()) == 1) {
    ans[elts[0]] = {xL, xH};
    return;
  }
  int targ = (int((elts).size()) + 1) / 2;
  ll vM = vL + (vH - vL) / int((elts).size()) * targ;
  vi left, middle = elts, right;
  ll lX = xL, rX = xH;
  while (true) {
    int pivot = middle[rng() % int((middle).size())];
    ll x = search(pivot, lX, rX, vM);
    vi before, equal = {pivot}, after;
    for (int i : middle)
      if (i != pivot) {
        ll v = eval(i, x);
        (v == vM ? equal : v > vM ? before : after).push_back(i);
      }
    if (int((left).size()) + int((before).size()) >= targ) {
      middle.clear();
      for (int i : before) middle.push_back(i);
      for (int i : equal) middle.push_back(i);
      for (int i : after) right.push_back(i);
      rX = x;
    } else if (int((left).size()) + int((before).size()) +
                   int((equal).size()) >=
               targ) {
      for (int i : before) left.push_back(i);
      for (int i : equal)
        (int((left).size()) < targ ? left : right).push_back(i);
      for (int i : after) right.push_back(i);
      solve(ans, left, xL, x, vL, vM);
      solve(ans, right, x, xH, vM, vH);
      return;
    } else {
      middle.clear();
      for (int i : before) left.push_back(i);
      for (int i : equal) middle.push_back(i);
      for (int i : after) middle.push_back(i);
      lX = x;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  int N;
  ll L;
  cin >> N >> L;
  vi elts(N);
  for (int i = 0; i < N; i++) elts[i] = i;
  vector<array<ll, 2>> ans(N);
  solve(ans, elts, 0, 1e18, 0, L);
  cout << "!\n";
  for (int i = 0; i < N; i++) cout << ans[i][0] << " " << ans[i][1] << "\n";
  cout.flush();
  return 0;
}