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 NMAX = 1010;
map<pair<int, long long>, long long> poz;
long long ans[2][NMAX];
mt19937 rnd(time(0));
long long L;
long long ask(int id, long long p) {
  if (poz.find({id, p}) != poz.end()) return poz[{id, p}];
  cout << "? " << id << ' ' << p << endl;
  cin >> poz[{id, p}];
  return poz[{id, p}];
}
long long interval(int x, long long a, long long b) {
  return ask(x, b) - ask(x, a);
}
long long find(int id, long long st, long long dr, long long S) {
  long long ans = st;
  for (long long i = (1LL << 60); i; i /= 2)
    if (ans + i <= dr && interval(id, st, ans + i) <= S) ans += i;
  assert(interval(id, st, ans) == S);
  return ans;
}
void kthelement(vector<int>& v, long long st, long long dr, long long S,
                int from, int to, int k) {
  if (from == to) return;
  shuffle(v.begin() + from, v.begin() + to + 1, rnd);
  int id = v[from];
  long long loc = find(id, st, dr, S);
  int last = from;
  for (int i = from + 1; i <= to; i++) {
    if (make_pair(interval(v[i], st, loc), i) > make_pair(S, id)) {
      swap(v[last], v[last + 1]);
      if (last != i - 1) swap(v[last], v[i]);
      last++;
    }
  }
  assert(v[last] == id);
  if (k < last)
    kthelement(v, st, dr, S, from, last - 1, k);
  else if (k > last)
    kthelement(v, st, dr, S, last + 1, to, k);
}
void solve(vector<int> v, long long st, long long dr) {
  while (st >= dr)
    ;
  int n = v.size();
  int m = n / 2;
  if (n == 1) {
    ans[0][v[0]] = st, ans[1][v[0]] = dr;
    return;
  }
  long long S = m * L;
  kthelement(v, st, dr, S, 0, v.size() - 1, m - 1);
  vector<int> left, right;
  for (int i = 0; i < v.size(); i++) {
    if (i < m)
      left.push_back(v[i]);
    else
      right.push_back(v[i]);
  }
  long long mij = find(left.back(), st, dr, S);
  assert(interval(left.back(), st, mij) == S);
  solve(left, st, mij);
  solve(right, mij, dr);
}
int main() {
  int n;
  cin >> n >> L;
  L /= n;
  vector<int> v(n);
  iota(v.begin(), v.end(), 1);
  long long MAX = 1000 * 1000 * 1000;
  MAX = MAX * MAX;
  solve(v, 0, MAX);
  cout << "!\n";
  for (int i = 1; i <= n; i++) cout << ans[0][i] << ' ' << ans[1][i] << '\n';
  cout.flush();
  return 0;
}