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;
class DebugStream {
} LOG;
template <typename T>
DebugStream &operator<<(DebugStream &s, const T &) {
  return s;
}
mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
const int maxN = 1e3 + 9, maxV = 1e6 + 9, MOD = 1e9 + 7, SQ = 335, lg = 20,
          bs = 29;
const long long MVV = 1e18;
pair<long long, long long> ans[maxN];
long long L, U;
int n, ind[maxN];
map<long long, long long> memo[maxN];
void assertion(bool x) {
  if (!x) exit(0);
}
int qcount;
long long query(int w, long long x) {
  if (memo[w].find(x) == memo[w].end()) {
    assertion(qcount < 200000);
    qcount++;
    cout << "? " << w + 1 << ' ' << x << '\n';
    cout.flush();
    long long resp;
    cin >> resp;
    memo[w][x] = resp;
  }
  return memo[w][x];
}
long long bins(int w, long long targ) {
  long long le = -1, ri = MVV, mi;
  while (le + 1 < ri) {
    mi = (le + ri) / 2;
    if (query(w, mi) >= targ)
      ri = mi;
    else
      le = mi;
  }
  return ri;
}
void solve(int *arr, int n, long long le, long long ri, int acu) {
  if (n == 1) return void(ans[*arr] = pair<long long, long long>(le, ri));
  int *one = arr, *two = arr + n, *mid;
  long long cr;
  while (one != arr + n / 2) {
    cr = bins(*(two - 1), (acu + n / 2) * U);
    mid = stable_partition(
        one, two, [&](int x) { return query(x, cr) >= (acu + n / 2) * U; });
    if (mid <= arr + n / 2) {
      one = mid;
      continue;
    }
    mid = stable_partition(one, mid, [&](int x) {
      return query(x, cr) > (acu + n / 2) * U ||
             query(x, cr - 1) != (acu + n / 2) * U - 1;
    });
    if (mid <= arr + n / 2)
      break;
    else
      two = mid;
  }
  solve(arr, n / 2, le, cr, acu);
  solve(arr + n / 2, n - n / 2, cr, ri, acu + n / 2);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> L;
  U = L / n;
  iota(ind, ind + n, 0);
  int *ptr = ind + n;
  shuffle(ind, ind + n, rnd);
  solve(ind, n, 0, MVV, 0);
  cout << "!\n";
  for (int i = 0; i < n; i++)
    cout << ans[i].first << ' ' << ans[i].second << '\n';
  LOG << n << ' ' << qcount << '\n';
}