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 rng;
ll MAX = 1e18;
const int MN = 1e3 + 100;
int get_rand(int x) {
  return static_cast<int>(static_cast<ll>(rng()) * x /
                          (static_cast<ll>(rng.max()) - rng.min() + 1));
}
int N;
ll L, E;
array<ll, 2> ans[MN];
ll get(int f, ll x) {
  printf("? %d %lld\n", f, x);
  fflush(stdout);
  ll r;
  scanf("%lld", &r);
  return r;
}
ll loc(int f, ll v, ll l = 0, ll r = MAX) {
  r++;
  ll m;
  while (r - l > 1) {
    m = l + (r - l >> 1);
    if (get(f, m) > v)
      r = m;
    else
      l = m;
  }
  return l;
}
void solve(vector<int>& v, int l, int r, ll lp, ll rp) {
  if (v.size() == 1) return ans[v[0]] = {lp, rp}, void();
  random_shuffle(v.begin(), v.end(), get_rand);
  int m = l + (r - l >> 1);
  ll mp;
  function<ll(vector<int>&, int)> kth_element = [&](vector<int>& x, int k) {
    vector<int> a, b, c;
    b.push_back(x.back());
    ll t = loc(x.back(), m * E, lp, rp);
    for (x.pop_back(); !x.empty(); x.pop_back()) {
      ll q = get(x.back(), t);
      if (m * E < q)
        a.push_back(x.back());
      else if (q < m * E)
        c.push_back(x.back());
      else
        b.push_back(x.back());
    }
    if (k < a.size())
      t = kth_element(a, k);
    else if (a.size() + b.size() <= k)
      t = kth_element(c, k - a.size() - b.size());
    for (int i : a) x.push_back(i);
    for (int i : b) x.push_back(i);
    for (int i : c) x.push_back(i);
    return t;
  };
  mp = kth_element(v, m - l);
  vector<int> L, R;
  for (int i = 0; i < m - l; i++) L.push_back(v[i]);
  for (int i = m - l; i < r - l; i++) R.push_back(v[i]);
  solve(L, l, m, lp, mp);
  solve(R, m, r, mp, rp);
}
int main(void) {
  srand(clock());
  scanf("%d%lld", &N, &L);
  E = L / N;
  vector<int> a;
  for (int i = 1; i <= N; i++) a.push_back(i);
  solve(a, 0, N, 0, MAX);
  printf("!\n");
  for (int i = 1; i <= N; i++) printf("%lld %lld\n", ans[i][0], ans[i][1]);
  fflush(stdout);
  return 0;
}