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;
int n;
long long L;
pair<long long, long long> ans[1005];
mt19937_64 rng(233);
long long F(int i, long long x) {
  printf("? %d %lld\n", i, x);
  fflush(stdout);
  long long y;
  scanf("%lld", &y);
  return y;
}
void sol(vector<pair<int, long long> > cur, long long l, long long r) {
  if (((int)(cur).size()) <= 1) {
    for (auto x : cur) {
      ans[x.first] = {l, r};
    }
    return;
  }
  vector<pair<int, long long> > can(cur);
  auto get = [&](int i, long long x) {
    return F(can[i].first, x) - can[i].second;
  };
  int bs = 0;
  vector<pair<int, long long> > left;
  vector<pair<int, long long> > right;
  int ned = ((int)(cur).size()) / 2;
  while (1) {
    int id = rng() % ((int)(can).size());
    long long lb = l, rb = r, res = -1;
    while (lb <= rb) {
      long long mid = (lb + rb) >> 1;
      if (get(id, mid) >= ned * (L / n)) {
        res = mid;
        rb = mid - 1;
      } else {
        lb = mid + 1;
      }
    }
    int greater = bs, equal = 0;
    vector<long long> w(((int)(can).size()));
    for (int i = (0); i <= (((int)(can).size()) - 1); ++i) {
      w[i] = get(i, res);
      if (w[i] > ned * (L / n)) {
        ++greater;
      } else if (w[i] == ned * (L / n)) {
        ++equal;
      }
    }
    if (greater + 1 <= ned && ned <= greater + equal) {
      vector<bool> bo(((int)(can).size()));
      for (int i = (0); i <= (((int)(can).size()) - 1); ++i) {
        if (w[i] > ned * (L / n)) {
          left.push_back(can[i]);
          bo[i] = 1;
        }
      }
      for (int i = (0); i <= (((int)(can).size()) - 1); ++i)
        if (!bo[i]) {
          if (w[i] >= ned * (L / n) && ((int)(left).size()) < ned) {
            left.push_back(can[i]);
          } else {
            right.push_back(can[i]);
          }
        }
      for (auto& x : right) {
        x.second = F(x.first, res);
      }
      sol(left, l, res);
      sol(right, res, r);
      return;
    }
    if (greater + 1 > ned) {
      vector<pair<int, long long> > ncan;
      for (int i = (0); i <= (((int)(can).size()) - 1); ++i) {
        if (w[i] < ned * (L / n)) {
          right.push_back(can[i]);
        } else
          ncan.push_back(can[i]);
      }
      can = ncan;
    }
    if (ned > greater + equal) {
      vector<pair<int, long long> > ncan;
      for (int i = (0); i <= (((int)(can).size()) - 1); ++i) {
        if (w[i] >= ned * (L / n)) {
          left.push_back(can[i]);
        } else
          ncan.push_back(can[i]);
      }
      can = ncan;
      bs = greater + equal;
    }
  }
}
int main() {
  scanf("%d%lld", &n, &L);
  vector<pair<int, long long> > tp(n, {0, 0});
  for (int i = (0); i <= (n - 1); ++i) tp[i].first = i + 1;
  sol(tp, 0, 1e18);
  printf("!\n");
  for (int i = (1); i <= (n); ++i)
    printf("%lld %lld\n", ans[i].first, ans[i].second);
  return 0;
}