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 = 1e3 + 10;
const long long X = 1e18;
int n;
long long L, l[N], r[N], F[N][N], lst;
long long f(int i, long long x) {
  cout << '?' << ' ' << i << ' ' << x << endl;
  long long y;
  cin >> y;
  return y;
}
long long bin(int i, long long l, long long r, long long v) {
  while (l < r) {
    long long mid = (l + r) >> 1;
    if (f(i, mid) < v)
      l = mid + 1;
    else
      r = mid;
  }
  return l;
}
vector<int> kth(vector<int> q, long long lx, long long rx, long long y, int k) {
  int i = q[rand() % int(q.size())];
  long long x = bin(i, lx, rx, y);
  assert(f(i, lx) <= y && y <= f(i, rx));
  assert(f(i, x) == y);
  vector<int> ls, rs, now;
  for (auto i : q) {
    long long v = f(i, x);
    if (v == y)
      now.push_back(i);
    else if (v > y)
      ls.push_back(i);
    else
      rs.push_back(i);
  }
  vector<int> ret;
  if (int(ls.size()) >= k) {
    assert(int(ls.size()) < int(q.size()));
    ret = kth(ls, lx, rx, y, k);
    ret.insert(ret.end(), now.begin(), now.end());
    ret.insert(ret.end(), rs.begin(), rs.end());
  } else if (int(ls.size()) + int(now.size()) >= k) {
    lst = x;
    ret = ls;
    ret.insert(ret.end(), now.begin(), now.end());
    ret.insert(ret.end(), rs.begin(), rs.end());
  } else {
    assert(int(rs.size()) < int(q.size()));
    ret = ls;
    ret.insert(ret.end(), now.begin(), now.end());
    rs = kth(rs, lx, rx, y, k - int(ls.size()) - int(now.size()));
    ret.insert(ret.end(), rs.begin(), rs.end());
  }
  return ret;
}
void solve(vector<int> q, long long lx, long long rx, int o) {
  if (int(q.size()) == 1) {
    l[q[0]] = lx;
    r[q[0]] = rx;
    return;
  }
  int mid = int(q.size()) / 2;
  lst = -1;
  q = kth(q, lx, rx, L / n * (o + mid), mid);
  assert(lst >= 0);
  long long x = lst;
  solve(vector<int>(q.begin(), q.begin() + mid), lx, x, o);
  solve(vector<int>(q.begin() + mid, q.end()), x, rx, o + mid);
}
int main() {
  srand(114514);
  cin >> n >> L;
  vector<int> q(n);
  for (int i = 0; i < n; i++) q[i] = i + 1;
  solve(q, 0, X, 0);
  for (int i = 1; i <= n; i++) assert(f(i, r[i]) - f(i, l[i]) >= L / n);
  puts("!");
  for (int i = 1; i <= n; i++) cout << l[i] << ' ' << r[i] << endl;
  return 0;
}