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;
using pll = pair<ll, ll>;
mt19937 rng(chrono::steady_clock().now().time_since_epoch().count());
int rand(int a, int b) { return uniform_int_distribution<int>(a, b)(rng); }
int que = 0;
ll ask(int func, ll x) {
  que++;
  if (que > 200000) exit(-1);
  printf("? %d %lld\n", func, x);
  fflush(stdout);
  ll ans;
  scanf("%lld", &ans);
  return ans;
}
ll find_first(int func, ll val) {
  ll pocz = 0, kon = 1e18, mid;
  while (pocz < kon) {
    mid = (pocz + kon) / 2;
    if (ask(func, mid) >= val)
      kon = mid;
    else
      pocz = mid + 1;
  }
  return pocz;
}
ll find_last(int func, ll val) {
  ll pocz = 0, kon = 1e18, mid;
  while (pocz < kon) {
    mid = (pocz + kon + 1) / 2;
    if (ask(func, mid) <= val)
      pocz = mid;
    else
      kon = mid - 1;
  }
  return pocz;
}
vector<int> find_k_smallest(vector<int> func, int k, ll val) {
  assert(k >= 0 && k <= func.size());
  if (k == 0 || func.empty()) return {};
  if (k == (int)func.size()) return func;
  swap(func[0], func[rand(0, func.size() - 1)]);
  ll where0 = find_first(func[0], val);
  vector<int> smaller, equal, larger;
  for (int i = 0; i < func.size(); i++) {
    ll hlp;
    if ((hlp = ask(func[i], where0)) > val)
      smaller.push_back(func[i]);
    else if (hlp < val)
      larger.push_back(func[i]);
    else
      equal.push_back(func[i]);
  }
  if (k <= smaller.size()) return find_k_smallest(smaller, k, val);
  if (k <= smaller.size() + equal.size()) {
    int ss = smaller.size();
    for (int i = 0; i < k - ss; i++) smaller.push_back(equal[i]);
    return smaller;
  }
  vector<int> hlp =
      find_k_smallest(larger, k - smaller.size() - equal.size(), val);
  for (int x : equal) smaller.push_back(x);
  for (int x : hlp) smaller.push_back(x);
  return smaller;
}
pll res[1007];
int n;
ll L;
int counter[1007];
int cnt;
void solve(vector<int> func, ll l1, ll l2) {
  if (func.empty()) return;
  ll l = L / n;
  assert(l2 >= l1);
  assert((l2 - l1) % l == 0);
  assert((l2 - l1) / l == func.size());
  if (func.size() == 1) {
    res[func[0]] = {0, find_first(func[0], l2)};
    return;
  }
  int n = func.size();
  int sn = n / 2;
  ll lmid = l1 + (l2 - l1) / n * sn;
  int qer = que;
  vector<int> fst = find_k_smallest(func, sn, lmid);
  solve(fst, l1, lmid);
  cnt++;
  for (int x : fst) counter[x] = cnt;
  vector<int> sc;
  for (int x : func)
    if (counter[x] != cnt) sc.push_back(x);
  solve(sc, lmid, l2);
}
void answer() {
  printf("!\n");
  vector<pll> hlp;
  for (int i = 1; i <= n; i++) hlp.emplace_back(res[i].second, i);
  sort(hlp.begin(), hlp.end());
  for (int i = 1; i < n; i++)
    res[hlp[i].second].first = res[hlp[i - 1].second].second;
  for (int i = 1; i <= n; i++) {
    printf("%lld %lld\n", res[i].first, res[i].second);
  }
}
int main() {
  scanf("%d%lld", &n, &L);
  vector<int> hlp;
  for (int i = 1; i <= n; i++) hlp.push_back(i);
  solve(hlp, 0, L);
  answer();
  return 0;
}