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 long long SEED = chrono::steady_clock::now().time_since_epoch().count();
mt19937_64 rng = mt19937_64(SEED);
inline long long rnd(const long long& a, const long long& b) {
  return uniform_int_distribution<long long>(a, b)(rng);
}
vector<pair<long long, long long> > ans;
long long n, L;
map<pair<int, long long>, long long> vv;
long long valueAt(int i, long long x) {
  if (vv.count({i, x}) == 0) {
    cout << "? " << i << " " << x << endl;
    long long res;
    cin >> res;
    vv[{i, x}] = res;
  }
  return vv[{i, x}];
}
long long getX(const int& i, long long l, long long h, long long v) {
  long long lg = 0;
  while (l <= h) {
    long long m = (l + h) / 2;
    long long x = valueAt(i, m);
    if (x < v) {
      lg = m;
      l = m + 1;
    } else {
      h = m - 1;
    }
  }
  return lg + 1;
}
long long splitPoint;
pair<vector<int>, vector<int> > splitInto(vector<int>& fun, long long lx,
                                          long long hx, long long ly,
                                          long long hy, long long my, int k) {
  int rPos = rnd(0, fun.size() - 1);
  long long x = getX(fun[rPos], lx, hx, my);
  vector<int> l, e, h;
  for (int i = 0; i < fun.size(); i++) {
    long long va = valueAt(fun[i], x);
    if (va < my) l.push_back(fun[i]);
    if (va == my) e.push_back(fun[i]);
    if (va > my) h.push_back(fun[i]);
  }
  while (e.size() > 0 && l.size() < k) {
    l.push_back(e.back());
    e.pop_back();
  }
  while (e.size() > 0) {
    h.push_back(e.back());
    e.pop_back();
  }
  if (l.size() > k) {
    auto pr = splitInto(l, x, hx, ly, hy, my, k);
    l = pr.first;
    for (auto a : pr.second) h.push_back(a);
  } else if (l.size() < k) {
    auto pr = splitInto(h, lx, x, ly, hy, my, k - l.size());
    h = pr.second;
    for (auto a : pr.first) l.push_back(a);
  } else
    splitPoint = x;
  return {l, h};
}
void solve(vector<int>& fun, long long lx, long long hx, long long ly,
           long long hy) {
  if (fun.size() == 0) return;
  if (fun.size() == 1) {
    ans[fun[0] - 1] = {lx, hx};
    return;
  }
  long long split = fun.size() / 2;
  long long my = ly + split * (L / n);
  pair<vector<int>, vector<int> > pr;
  if (fun.size() % 2 == 0)
    pr = splitInto(fun, lx, hx, ly, hy, my, split);
  else
    pr = splitInto(fun, lx, hx, ly, hy, my, split + 1);
  vector<int> fl = move(pr.second);
  vector<int> fr = move(pr.first);
  long long p = splitPoint;
  solve(fl, lx, p, ly, my);
  solve(fr, p, hx, my, hy);
}
int32_t main(int argv, char** argc) {
  ios::sync_with_stdio(false);
  cin >> n >> L;
  ans.resize(n);
  vector<int> fun;
  for (int i = 0; i < n; i++) fun.push_back(i + 1);
  shuffle(fun.begin(), fun.end(), rng);
  solve(fun, 0, 1e18, 0, L);
  cout << "!\n";
  for (auto a : ans) cout << a.first << " " << a.second << "\n";
  cout << endl;
}