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 long long Lim = 1e18;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
long long tot = 0;
unordered_map<long long, long long> mmp;
long long query(long long w, long long first) {
  if (mmp.count((w << 31) + first)) return mmp[(w << 31) + first];
  long long v;
  if (++tot == 2e5) exit(0);
  printf("? %lld %lld\n", w, first), fflush(stdout);
  scanf("%lld", &v);
  return mmp[(w << 31) + first] = v;
}
long long findx(long long w, long long first, long long L, long long R) {
  for (; L <= R;) {
    long long md = (L + R) / 2, v = query(w, md);
    if (v == first) return md;
    if (v < first)
      L = md + 1;
    else
      R = md - 1;
  }
}
pair<long long, long long> Ans[2010];
long long det;
void Solve(vector<int> V, long long rL, long long rR, long long vL,
           long long vR) {
  if (vL + 1 == vR) {
    assert(V.size() == 1);
    Ans[V[0]] = pair<long long, long long>(
        query(V[0], rL) == vL * det ? rL : findx(V[0], vL * det, rL, rR),
        query(V[0], rR) == vR * det ? rR : findx(V[0], vR * det, rL, rR));
    return;
  }
  long long md = (vL + vR) / 2, lsz = md - vL, rsz = vR - md;
  vector<int> LV(0), RV(0);
  for (;;) {
    int w = V[rng() % V.size()];
    long long o = findx(w, md * det, rL, rR);
    vector<int> tmpL(0), tmpR(0), tmpmd(0);
    for (int i = 0; i < V.size(); i++) {
      long long tmp = query(V[i], o);
      if (tmp < md * det)
        tmpR.push_back(V[i]);
      else if (tmp == md * det)
        tmpmd.push_back(V[i]);
      else
        tmpL.push_back(V[i]);
    }
    if ((int)tmpL.size() + (int)LV.size() > lsz) {
      RV.insert(RV.end(), tmpR.begin(), tmpR.end());
      RV.insert(RV.end(), tmpmd.begin(), tmpmd.end());
      V = tmpL;
    } else if ((int)tmpR.size() + (int)RV.size() > rsz) {
      LV.insert(LV.end(), tmpL.begin(), tmpL.end());
      LV.insert(LV.end(), tmpmd.begin(), tmpmd.end());
      V = tmpR;
    } else {
      LV.insert(LV.end(), tmpL.begin(), tmpL.end());
      RV.insert(RV.end(), tmpR.begin(), tmpR.end());
      LV.insert(LV.end(), tmpmd.begin(), tmpmd.begin() + (lsz - LV.size()));
      RV.insert(RV.end(), tmpmd.end() - (rsz - RV.size()), tmpmd.end());
      Solve(LV, rL, o, vL, md);
      Solve(RV, o, rR, md, vR);
      return;
    }
  }
}
int main() {
  long long n, L;
  scanf("%lld%lld", &n, &L), det = L / n;
  vector<int> V(n);
  for (int i = 0; i < n; i++) V[i] = i + 1;
  Solve(V, 0, Lim, 0, n);
  putchar('!');
  putchar('\n');
  for (int i = 1; i <= n; i++)
    printf("%lld %lld\n", Ans[i].first, Ans[i].second);
  fflush(stdout);
  return 0;
}