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;
long long n, L, q[1005], q1[1005], q2[1005], q3[1005], q1_cnt, q2_cnt, q3_cnt,
    ansl[1005], ansr[1005], rec;
void work(long long ql, long long qr, long long level, long long tar,
          long long nl, long long nr) {
  if (ql > qr) return;
  q1_cnt = q2_cnt = q3_cnt = 0;
  long long w = ql + rand() % (qr - ql + 1);
  long long l = nl, r = nr;
  while (l <= r) {
    long long mid = l + r >> 1;
    cout << '?' << ' ' << q[w] << ' ' << mid << endl;
    long long val;
    cin >> val;
    if (val >= level)
      r = mid - 1;
    else
      l = mid + 1;
  }
  rec = l;
  q2[++q2_cnt] = q[w];
  for (long long i = ql; i <= qr; i++)
    if (i != w) {
      cout << '?' << ' ' << q[i] << ' ' << l << endl;
      long long val;
      cin >> val;
      if (val < level)
        q3[++q3_cnt] = q[i];
      else if (val > level)
        q1[++q1_cnt] = q[i];
      else
        q2[++q2_cnt] = q[i];
    }
  for (long long i = 1; i <= q1_cnt; i++) q[ql + i - 1] = q1[i];
  for (long long i = 1; i <= q2_cnt; i++) q[ql + q1_cnt + i - 1] = q2[i];
  for (long long i = 1; i <= q3_cnt; i++)
    q[ql + q1_cnt + q2_cnt + i - 1] = q3[i];
  if (tar >= ql + q1_cnt && tar < ql + q1_cnt + q2_cnt) return;
  if (tar < ql + q1_cnt)
    work(ql, ql + q1_cnt - 1, level, tar, nl, nr);
  else
    work(ql + q1_cnt + 1, qr, level, tar, nl, nr);
}
void solve(long long nl, long long nr, long long low, long long top,
           long long ql, long long qr) {
  if (ql == qr) {
    ansl[q[ql]] = nl;
    ansr[q[qr]] = nr;
    return;
  }
  long long cnt = qr - ql + 1 >> 1, level = low + L / n * cnt;
  work(ql, qr, level, ql + cnt - 1, nl, nr);
  long long tmp = rec;
  solve(nl, tmp, low, level, ql, ql + cnt - 1);
  solve(tmp, nr, level, top, ql + cnt, qr);
}
signed main() {
  srand(time(NULL));
  cin >> n >> L;
  for (long long i = 1; i <= n; i++) q[i] = i;
  solve(0, 1e18, 0, L, 1, n);
  cout << '!' << '\n';
  for (long long i = 1; i <= n; i++) cout << ansl[i] << ' ' << ansr[i] << '\n';
  return 0;
}