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;
template <class TH>
void _dbg(const char *sdbg, TH h) {
  cerr << sdbg << '=' << h << endl;
}
template <class TH, class... TA>
void _dbg(const char *sdbg, TH h, TA... a) {
  while (*sdbg != ',') cerr << *sdbg++;
  cerr << '=' << h << ',';
  _dbg(sdbg + 1, a...);
}
template <class T>
ostream &operator<<(ostream &os, vector<T> V) {
  os << "[";
  for (auto vv : V) os << vv << ", ";
  return os << "]";
}
template <class L, class R>
ostream &operator<<(ostream &os, pair<L, R> P) {
  return os << "(" << P.first << ", " << P.second << ")";
}
long long n, last;
long long tab[1123][1123];
map<pair<long long, long long>, long long> gettab;
long long get(long long i, long long x) {
  if (gettab.find(pair<long long, long long>(i, x)) != gettab.end())
    return gettab[pair<long long, long long>(i, x)];
  long long retv = x;
  cout << "? " << i << " " << x << endl;
  cin >> retv;
  return gettab[pair<long long, long long>(i, x)] = retv;
}
pair<long long, long long> ans[1123];
int pick(int a, int b) { return a + (rand() % (b - a + 1)); }
long long find_fmid(int i, long long l, long long r, long long val) {
  while (l != r) {
    long long mid = (l + r) / 2;
    if (get(i, mid) >= val)
      r = mid;
    else
      l = mid + 1;
  }
  return l;
}
long long split(long long l, long long r, long long prev, vector<int> pi,
                vector<int> &lpi, vector<int> &rpi) {
  int a = 0, b = pi.size() - 1;
  int k = pi.size() / 2;
  long long xp;
  while (a != b) {
    assert(a <= k && k <= b);
    int pivot = pick(a, b);
    xp = find_fmid(pi[pivot], l, r, (prev + k) * (last / n));
    lpi.clear();
    rpi.clear();
    vector<int> eq;
    for (int i = a; i <= b; i++)
      if (i != pivot) {
        if (get(pi[i], xp) > (prev + k) * (last / n))
          lpi.push_back(pi[i]);
        else if (get(pi[i], xp) < (prev + k) * (last / n))
          rpi.push_back(pi[i]);
        else
          eq.push_back(pi[i]);
      }
    int bla = pi[pivot];
    int tmp = a;
    for (int i = 0; i < lpi.size(); i++) pi[tmp++] = lpi[i];
    for (int i = 0; i < eq.size(); i++) pi[tmp++] = eq[i];
    int breakpoint = tmp;
    pi[tmp++] = bla;
    for (int i = 0; i < rpi.size(); i++) pi[tmp++] = rpi[i];
    if (k > breakpoint) {
      a = breakpoint + 1;
    } else if (k < breakpoint - eq.size()) {
      b = breakpoint - 1;
    } else {
      swap(pi[k], pi[breakpoint]);
      break;
    }
  }
  lpi.clear();
  rpi.clear();
  for (int i = 0; i < k; i++) lpi.push_back(pi[i]);
  for (int i = k; i < pi.size(); i++) rpi.push_back(pi[i]);
  return find_fmid(pi[k], l, r, (prev + k) * (last / n));
}
void solve(long long l, long long r, long long prev, vector<int> pi) {
  if (pi.size() == 1) {
    ans[pi[0]] = {l, r};
    return;
  }
  vector<int> lpi, rpi;
  long long mid = split(l, r, prev, pi, lpi, rpi);
  solve(l, mid, prev, lpi);
  solve(mid, r, prev + lpi.size(), rpi);
}
int main(void) {
  cin >> n >> last;
  while (true) {
    vector<int> pi;
    for (int i = 1; i <= n; i++) pi.push_back(i);
    solve(0, 1e18, 0, pi);
    cout << "!" << endl;
    for (int i = 1; i <= n; i++) {
      cout << ans[i].first << " " << ans[i].second << endl;
    }
    break;
  }
}