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 mxN = 1e3;
int n, a[mxN], b[mxN];
long long L, l[mxN], r[mxN];
long long ask(int i, long long x) {
  cout << "? " << i + 1 << " " << x << endl;
  cin >> x;
  return x;
}
void dc(int l1, int r1, long long l2, long long r2) {
  if (l1 == r1) {
    l[a[l1]] = l2;
    r[a[l1]] = r2;
    return;
  }
  int m1 = (l1 + r1) / 2;
  long long x = L / n * (m1 + 1), m2;
  for (int l = l1, r = r1;;) {
    int i = rand() % (r - l + 1) + l;
    long long lb = l2, rb = r2;
    while (lb < rb) {
      long long mb = (lb + rb) / 2;
      if (ask(a[i], mb) < x)
        lb = mb + 1;
      else
        rb = mb;
    }
    int l2 = l, r2 = r;
    vector<int> e;
    for (int j = l; j <= r; ++j) {
      long long y = ask(a[j], lb);
      if (y > x)
        b[l2++] = a[j];
      else if (y < x)
        b[r2--] = a[j];
      else
        e.push_back(a[j]);
    }
    for (int j = 0; j < e.size(); ++j) b[l2 + j] = e[j];
    memcpy(a + l, b + l, 4 * (r - l + 1));
    if (l2 <= m1 && m1 <= r2) {
      m2 = lb;
      break;
    }
    if (l2 < m1)
      l = l2 + 1;
    else
      r = l2 - 1;
  }
  dc(l1, m1, l2, m2);
  dc(m1 + 1, r1, m2, r2);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> L;
  iota(a, a + n, 0);
  dc(0, n - 1, 0, 1e18);
  cout << "!\n";
  for (int i = 0; i < n; ++i) cout << l[i] << " " << r[i] << "\n";
}