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 ll = long long;
const int N = 1010;
const ll MAX = (ll)1e18;
int n;
int rand(int n) { return rand() % n; }
ll ask(int id, ll pos) {
  printf("? %d %lld\n", id + 1, pos);
  fflush(stdout);
  ll ret;
  scanf("%lld", &ret);
  return ret;
}
ll left[N], right[N];
void solve(std::vector<int> vec, ll l, ll r, ll l_value, ll r_value) {
  if ((int)vec.size() == 1) {
    left[vec[0]] = l;
    right[vec[0]] = r;
    return;
  }
  ll diff = (r_value - l_value) / vec.size();
  ll target;
  if (vec.size() % 2 == 0) {
    target = (l_value + r_value) / 2;
  } else {
    target = (l_value + r_value - diff) / 2;
  }
  int target_sit = ((int)vec.size() + 1) / 2;
  std::vector<int> small, big, possible = vec;
  ll cur_l = l, cur_r = r;
  while (true) {
    int pos = possible[rand(possible.size())];
    std::vector<std::pair<ll, int>> value;
    ll l1 = cur_l, r1 = cur_r;
    while (l1 < r1) {
      ll mid = (l1 + r1) / 2;
      if (ask(pos, mid) >= target) {
        r1 = mid;
      } else {
        l1 = mid + 1;
      }
    }
    for (auto u : possible) {
      value.emplace_back(ask(u, l1), u);
    }
    std::sort(value.begin(), value.end());
    int sit = target_sit - (int)small.size();
    if (value[sit].first < target) {
      std::vector<int> npossible;
      for (auto p : value) {
        if (p.first >= target) {
          big.push_back(p.second);
        } else {
          npossible.push_back(p.second);
        }
      }
      possible = npossible;
      cur_l = l1 + 1;
    } else if (value[sit].first > target) {
      std::vector<int> npossible;
      for (auto p : value) {
        if (p.first <= target) {
          small.push_back(p.second);
        } else {
          npossible.push_back(p.second);
        }
      }
      possible = npossible;
      cur_r = l1 - 1;
    } else {
      for (int i = 0; i < sit; ++i) {
        small.push_back(value[i].second);
      }
      for (int i = sit; i < (int)value.size(); ++i) {
        big.push_back(value[i].second);
      }
      solve(small, l1, r, target, r_value);
      solve(big, l, l1, l_value, target);
      break;
    }
  }
}
int main() {
  srand((unsigned)time(nullptr));
  ll l;
  scanf("%d%lld", &n, &l);
  std::vector<int> vec(n);
  for (int i = 0; i < n; ++i) {
    vec[i] = i;
  }
  solve(vec, 0, MAX, 0, l);
  puts("!");
  for (int i = 0; i < n; ++i) {
    printf("%lld %lld\n", left[i], right[i]);
  }
  return 0;
}