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 bool TEST = false;
using ll = long long;
const ll MAXX = ll(1e18);
int n;
ll l;
ll D;
int nq = 0;
map<pair<int, ll>, ll> cache;
ll f(int i, ll x) {
  if (cache.count({i, x})) return cache[{i, x}];
  nq += 1;
  assert(nq <= 99999);
  if (TEST) return min(x / (i + 1), 5ll);
  assert(0 <= i && i < n);
  assert(0 <= x && x <= MAXX);
  cout << "?"
       << " " << i + 1 << " " << x << '\n';
  cout << flush;
  ll val;
  cin >> val;
  cache[{i, x}] = val;
  return val;
}
vector<pair<ll, ll> > ans;
mt19937 mt(48);
void solve(ll s, ll e, vector<int> ids) {
  int n = ids.size();
  if (n == 1) {
    ans[ids.front()] = {s, e};
    return;
  }
  int z = n / 2;
  vector<int> lids;
  vector<int> rids;
  while (!ids.empty()) {
    shuffle(ids.begin(), ids.end(), mt);
    int x = ids.front();
    ll l = s - 1;
    ll r = e + 1;
    ll loc = -1;
    assert(f(x, e) - f(x, s) >= n * D);
    while (l + 1 < r) {
      ll m = (l + r) / 2;
      bool lok = f(x, m) - f(x, s) >= z * D;
      bool rok = f(x, e) - f(x, m) >= (n - z) * D;
      if (lok && rok) {
        loc = m;
        break;
      } else if (rok) {
        l = m;
      } else if (lok) {
        r = m;
      } else
        assert(false);
    }
    assert(loc != -1);
    vector<int> lt, eq, rt;
    for (int x : ids) {
      assert(f(x, e) - f(x, s) >= n * D);
      bool lok = f(x, loc) - f(x, s) >= z * D;
      bool rok = f(x, e) - f(x, loc) >= (n - z) * D;
      if (lok && rok) {
        eq.push_back(x);
      } else if (lok) {
        lt.push_back(x);
      } else if (rok) {
        rt.push_back(x);
      } else
        assert(false);
    }
    ids.clear();
    if (lids.size() + lt.size() <= z) {
      for (int x : lt) lids.push_back(x);
      lt.clear();
      while (!eq.empty() && (int)lids.size() < z) {
        lids.push_back(eq.back());
        eq.pop_back();
      }
    }
    if (rids.size() + rt.size() <= n - z) {
      for (int x : rt) rids.push_back(x);
      rt.clear();
      while (!eq.empty() && (int)rids.size() < n - z) {
        rids.push_back(eq.back());
        eq.pop_back();
      }
    }
    for (int x : lt) ids.push_back(x);
    for (int x : eq) ids.push_back(x);
    for (int x : rt) ids.push_back(x);
    if (ids.empty()) {
      solve(s, loc, lids);
      solve(loc, e, rids);
      return;
    }
  }
  assert(false);
  return;
}
void answer() {
  cout << "!" << '\n';
  for (pair<ll, ll> x : ans) {
    assert(x.first != -1 && x.second != -1);
    cout << x.first << ' ' << x.second << '\n';
  }
  cout << flush;
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n >> l;
  D = l / n;
  ans.assign(n, {-1, -1});
  vector<int> ids;
  for (int i = 0; i < n; i++) ids.push_back(i);
  solve(0, MAXX, ids);
  answer();
}