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 = 1234;
const long long MAX = (long long)1e18;
mt19937 mrand(chrono::high_resolution_clock::now().time_since_epoch().count());
pair<long long, long long> intervals[N];
long long m, y;
int n;
long long query(int x, long long y) {
  cout << "? " << x + 1 << " " << y << endl;
  long long p;
  cin >> p;
  return p;
}
long long binary_search(int who, long long w, long long l, long long r) {
  while (true) {
    long long m = (l + r) >> 1;
    long long v = query(who, m);
    if (v > w) {
      r = m - 1;
    } else if (v < w) {
      l = m + 1;
    } else {
      return m;
    }
  }
}
pair<vector<int>, vector<int>> split_leftmost_k(vector<int> a, int o, int k,
                                                long long l, long long r) {
  vector<int> left, middle, right;
  int p = a[mrand() % a.size()];
  long long q = binary_search(p, m / n * o, l, r);
  middle.push_back(p);
  for (auto i : a) {
    if (i != p) {
      long long v = query(i, q);
      if (v > m / n * o) {
        left.push_back(i);
      } else if (v < m / n * o) {
        right.push_back(i);
      } else {
        middle.push_back(i);
      }
    }
  }
  while ((int)left.size() < k && !middle.empty()) {
    left.push_back(middle.back());
    middle.pop_back();
  }
  if ((int)left.size() == k) {
    y = q;
    for (auto x : middle) {
      right.push_back(x);
    }
    return make_pair(left, right);
  } else if ((int)left.size() < k) {
    pair<vector<int>, vector<int>> down =
        split_leftmost_k(right, o, k - left.size(), l, r);
    for (auto x : left) {
      down.first.push_back(x);
    }
    return down;
  } else {
    pair<vector<int>, vector<int>> down = split_leftmost_k(left, o, k, l, r);
    for (auto x : right) {
      down.second.push_back(x);
    }
    for (auto x : middle) {
      down.second.push_back(x);
    }
    return down;
  }
}
void solve(vector<int> a, int o, long long l, long long r) {
  if ((int)a.size() == 1) {
    intervals[a.back()] = make_pair(l, r);
  } else {
    int m = a.size() / 2;
    pair<vector<int>, vector<int>> left_and_right =
        split_leftmost_k(a, o + m, m, l, r);
    long long y = ::y;
    solve(left_and_right.first, o, l, y);
    solve(left_and_right.second, o + m, y, r);
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> m;
  vector<int> a(n);
  for (int i = 0; i < n; ++i) {
    a[i] = i;
  }
  solve(a, 0, 0, MAX);
  cout << "!" << endl;
  for (int i = 0; i < n; ++i) {
    cout << intervals[i].first << " " << intervals[i].second << endl;
  }
  return 0;
}