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 = 1005;
const long long INF = 1E18;
mt19937_64 mt(133769420);
vector<int> cl, ce, cr;
int n, a[N], st[N];
long long lim, al[N], ar[N];
int random(int n) {
  uniform_int_distribution<int> dis(0, n - 1);
  return dis(mt);
}
long long ask(int u, long long v) {
  cout << "? " << u << " " << v << endl;
  long long ret;
  cin >> ret;
  return ret;
}
long long BS(int u, long long tar, long long le, long long ri) {
  while (le <= ri) {
    long long mi = (le + ri) / 2;
    long long cur = ask(u, mi);
    if (cur == tar)
      return mi;
    else if (cur < tar)
      le = mi + 1;
    else
      ri = mi - 1;
  }
}
void solve(long long le, long long ri, int add, vector<int> ve) {
  int n = ve.size();
  if (n == 1) {
    al[ve.back()] = le;
    ar[ve.back()] = ri;
    return;
  }
  vector<int> ml, mr;
  long long pos, mid = lim * (add + n / 2);
  while (!ve.empty()) {
    int cl = ml.size(), cr = mr.size();
    pos = BS(ve[random(ve.size())], mid, le, ri);
    for (int i = 0; i < ve.size(); i++) {
      long long cur = ask(ve[i], pos);
      if (cur > mid)
        st[i] = 0;
      else if (cur == mid)
        st[i] = 1;
      else
        st[i] = 2;
      cl += (st[i] == 0);
      cr += (st[i] == 2);
    }
    for (int i = ve.size() - 1; i >= 0; i--) {
      if (st[i] == 0 && cl <= n / 2) {
        ml.push_back(ve[i]);
        ve[i] = ve.back();
        ve.pop_back();
      } else if (st[i] == 2 && cr <= n - n / 2) {
        mr.push_back(ve[i]);
        ve[i] = ve.back();
        ve.pop_back();
      } else if (st[i] == 1 && cl <= n / 2 && cr <= n - n / 2) {
        if (cl < n / 2) {
          ml.push_back(ve[i]);
          cl++;
        } else {
          mr.push_back(ve[i]);
          cr++;
        }
        ve[i] = ve.back();
        ve.pop_back();
      }
    }
  }
  solve(le, pos, add, ml);
  solve(pos, ri, add + n / 2, mr);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> n >> lim;
  lim /= n;
  vector<int> ve(n);
  for (int i = 0; i < n; i++) ve[i] = i + 1;
  solve(0, INF, 0, ve);
  cout << "!" << endl;
  for (int i = 1; i <= n; i++) cout << al[i] << " " << ar[i] << endl;
}