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;
mt19937 mt(48);
const int MAXN = 1100;
int N;
ll L;
ll lind[MAXN];
ll lval[MAXN];
ll rind[MAXN];
ll rval[MAXN];
map<pair<int, ll>, ll> cache;
int numQueries = 0;
ll query(int i, ll x) {
  pair<int, ll> st(i, x);
  if (cache.count(st)) return cache[st];
  if (x == 0) return cache[st] = 0;
  if (x == ll(1e18)) return cache[st] = L;
  numQueries++;
  cout << "?" << ' ' << i + 1 << ' ' << x << '\n' << flush;
  ll v;
  cin >> v;
  return cache[st] = v;
}
void dfs(vector<int> inds, ll l, ll r) {
  assert(!inds.empty());
  if (inds.size() == 1) {
    lind[inds[0]] = l;
    rind[inds[0]] = r;
    return;
  }
  assert(inds.size() >= 2);
  int ltarget = int(inds.size()) / 2;
  int rtarget = int(inds.size()) - ltarget;
  shuffle(inds.begin(), inds.end(), mt);
  vector<int> linds;
  vector<int> rinds;
  vector<int> minds = std::move(inds);
  ll curL = l, curR = r;
  while (true) {
    ll mi = 0, ma = ll(1e18);
    ll md;
    while (ma - mi > 1) {
      md = (mi + ma) / 2;
      if (md >= curR) {
        ma = md;
      } else if (md <= curL) {
        mi = md;
      } else {
        int i = minds[0];
        ll ldiff = query(i, md) - query(i, l);
        ll rdiff = query(i, r) - query(i, md);
        if (ldiff < ((L / N) * ltarget)) {
          mi = md;
        } else if (rdiff < ((L / N) * rtarget)) {
          ma = md;
        } else {
          break;
        }
      }
    }
    vector<int> lgoods;
    vector<int> rgoods;
    vector<int> bgoods;
    for (int i : minds) {
      ll ldiff = query(i, md) - query(i, l);
      ll rdiff = query(i, r) - query(i, md);
      bool lgood = ldiff >= ((L / N) * ltarget);
      bool rgood = rdiff >= ((L / N) * rtarget);
      assert(lgood || rgood);
      if (lgood && rgood) {
        bgoods.push_back(i);
      } else if (lgood) {
        lgoods.push_back(i);
      } else if (rgood) {
        rgoods.push_back(i);
      } else
        assert(false);
    }
    if (int(linds.size() + lgoods.size() + bgoods.size()) >= ltarget &&
        int(rinds.size() + rgoods.size() + bgoods.size()) >= rtarget) {
      linds.insert(linds.end(), lgoods.begin(), lgoods.end());
      rinds.insert(rinds.end(), rgoods.begin(), rgoods.end());
      while (int(linds.size()) < ltarget) {
        linds.push_back(bgoods.back());
        bgoods.pop_back();
      }
      rinds.insert(rinds.end(), bgoods.begin(), bgoods.end());
      dfs(linds, l, md);
      dfs(rinds, md, r);
      return;
    } else if (int(linds.size() + lgoods.size() + bgoods.size()) < ltarget) {
      linds.insert(linds.end(), lgoods.begin(), lgoods.end());
      linds.insert(linds.end(), bgoods.begin(), bgoods.end());
      minds = std::move(rgoods);
      curL = md;
    } else {
      rinds.insert(rinds.end(), rgoods.begin(), rgoods.end());
      rinds.insert(rinds.end(), bgoods.begin(), bgoods.end());
      minds = std::move(lgoods);
      curR = md;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> N >> L;
  for (int i = 0; i < N; i++) {
    lind[i] = 0;
    lval[i] = 0;
    rind[i] = ll(1e18);
    rval[i] = L;
  }
  vector<int> inds;
  for (int i = 0; i < N; i++) inds.push_back(i);
  dfs(inds, 0, ll(1e18));
  cout << "!" << '\n';
  for (int i = 0; i < N; i++) {
    cout << lind[i] << ' ' << rind[i] << '\n';
  }
  cerr << "numQueries " << numQueries << '\n';
  return 0;
}