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 resl[1000], resr[1000];
map<pair<int, long long>, long long> cache;
long long query(int f, long long x) {
  if (cache.count({f, x})) return cache[{f, x}];
  cout << "? " << f << " " << x << "\n" << flush;
  long long r;
  cin >> r;
  cache[{f, x}] = r;
  return r;
}
long long findx(int f, long long a, long long b, long long m) {
  while (a < b) {
    long long c = (a + b) / 2;
    long long r = query(f, c);
    if (r < m)
      a = c + 1;
    else
      b = c;
  }
  return a;
}
vector<int> kth(vector<int> f, int k, long long a, long long b, long long &c,
                long long m) {
  swap(f[0], f[rand() % f.size()]);
  vector<int> x, eq, y;
  long long p = findx(f[0], a, b, m);
  eq.push_back(f[0]);
  for (int i = 1; i < f.size(); ++i) {
    long long r = query(f[i], p);
    if (r > m)
      x.push_back(f[i]);
    else if (r < m)
      y.push_back(f[i]);
    else
      eq.push_back(f[i]);
  }
  if (x.size() <= k && k < x.size() + eq.size())
    c = p;
  else if (k < x.size())
    x = kth(x, k, a, p - 1, c, m);
  else
    y = kth(y, k - x.size() - eq.size(), p + 1, b, c, m);
  for (int yy : eq) x.push_back(yy);
  for (int yy : y) x.push_back(yy);
  return x;
}
void calc(vector<int> f, long long a, long long b, long long l, long long s) {
  if (f.size() == 0) return;
  if (f.size() == 1) {
    resl[f[0] - 1] = a;
    resr[f[0] - 1] = b;
    return;
  }
  long long m = l + s * (f.size() / 2);
  long long c;
  f = kth(f, f.size() / 2, a, b, c, m);
  vector<int> x, y;
  for (int i = 0; i < f.size() / 2; ++i) x.push_back(f[i]);
  for (int i = f.size() / 2; i < f.size(); ++i) y.push_back(f[i]);
  calc(x, a, c, l, s);
  calc(y, c, b, m, s);
}
int main() {
  int n;
  long long L;
  cin >> n >> L;
  vector<int> f;
  for (int i = 1; i <= n; ++i) f.push_back(i);
  calc(f, 0, 1e18, 0, L / n);
  cout << "!\n";
  for (int i = 0; i < n; ++i) cout << resl[i] << " " << resr[i] << "\n";
}