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;
using ll = long long;
const ll H = 1e18;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template <class T>
T rand(T a, T b) {
  return uniform_int_distribution<T>(a, b)(rng);
}
map<pair<int, ll>, ll> memo;
ll ask(int i, ll x) {
  pair<int, ll> pr = {i, x};
  if (memo.find(pr) != memo.end()) return memo[pr];
  cout << "? " << i + 1 << ' ' << x << endl;
  ll r;
  cin >> r;
  memo[pr] = r;
  return r;
}
ll getInd(int i, ll t) {
  ll low = 0;
  ll high = H;
  while (low != high) {
    ll mid = (low + high) >> 1;
    if (ask(i, mid) >= t)
      high = mid;
    else
      low = mid + 1;
  }
  return low;
}
vector<int> recSort(ll t, vector<int> act, int orig_k = 0) {
  int n = act.size();
  if (n == 1) return act;
  int le_tc = (n + 1) / 2;
  int ri_tc = n / 2;
  int k = orig_k + le_tc;
  ll v = k * t;
  vector<int> le, ri;
  while (!act.empty()) {
    int m = act.size();
    int i = rand(0, m - 1);
    ll x = getInd(act[i], v);
    vector<ll> offs(m, 0);
    offs[i] = 0;
    int lesses = 0;
    int mores = 0;
    for (int j = 0; j < m; ++j) {
      if (j == i) continue;
      ll cou = ask(act[j], x - 1);
      if (cou >= v)
        offs[j] = -1;
      else if (cou < v - 1)
        offs[j] = 1;
      else {
        ll cou2 = ask(act[j], x);
        if (cou2 == v - 1) offs[j] = 1;
      }
      if (offs[j] < 0) ++lesses;
      if (offs[j] > 0) ++mores;
    }
    if (le.size() + lesses <= le_tc && ri.size() + mores <= ri_tc) {
      int add_le = le_tc - lesses - le.size();
      for (int j = 0; j < m; ++j) {
        if (offs[j] == -1)
          le.push_back(act[j]);
        else if (offs[j] == 1)
          ri.push_back(act[j]);
        else {
          if (add_le > 0) {
            le.push_back(act[j]);
            --add_le;
          } else {
            ri.push_back(act[j]);
          }
        }
      }
      act.clear();
    } else if (le.size() + lesses <= le_tc) {
      vector<int> nxt_act;
      for (int j = 0; j < m; ++j) {
        if (offs[j] <= 0)
          le.push_back(act[j]);
        else
          nxt_act.push_back(act[j]);
      }
      act = nxt_act;
    } else {
      vector<int> nxt_act;
      for (int j = 0; j < m; ++j) {
        if (offs[j] >= 0)
          ri.push_back(act[j]);
        else
          nxt_act.push_back(act[j]);
      }
      act = nxt_act;
    }
  }
  vector<int> le_sorted = recSort(t, le, orig_k);
  vector<int> ri_sorted = recSort(t, ri, orig_k + le_tc);
  vector<int> res;
  for (auto i : le_sorted) res.push_back(i);
  for (auto i : ri_sorted) res.push_back(i);
  return res;
}
int main() {
  int n;
  ll v;
  cin >> n >> v;
  ll t = v / n;
  vector<int> inds(n);
  for (int i = 0; i < n; ++i) inds[i] = i;
  vector<int> ord = recSort(t, inds);
  ll pre = 0;
  vector<pair<ll, ll>> ints(n);
  for (int j = 0; j < n; ++j) {
    int i = ord[j];
    ll x = getInd(i, t * (j + 1));
    ints[i] = {pre, x};
    pre = x;
  }
  cout << "!" << endl;
  for (int i = 0; i < n; ++i) {
    cout << ints[i].first << ' ' << ints[i].second << endl;
  }
}