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 int N = 1e3 + 5;
vector<int> operator+(vector<int> a, const vector<int>& b) {
  for (auto x : b) {
    a.push_back(x);
  }
  return a;
}
class rnd {
 public:
  rnd() { srand((unsigned)time(NULL)); }
  int rand(int l, int r) { return ::rand() % (r - l) + l; }
} rnd;
int n;
long long L, m;
pair<long long, long long> ans[N];
long long query(int idx, long long x) {
  long long rt;
  printf("? %d %lld\n", idx, x);
  fflush(stdout);
  scanf("%lld", &rt);
  return rt;
}
long long _f(int idx, long long y, long long l = 0, long long r = 1e18) {
  long long mid, qy;
  while (true) {
    mid = (l + r) >> 1;
    qy = query(idx, mid);
    if (qy < y) {
      l = mid + 1;
    } else if (qy > y) {
      r = mid - 1;
    } else {
      return mid;
    }
  }
  assert(false);
  return -1;
}
pair<vector<int>, vector<int> > kth_element(vector<int> a, int k, long long v,
                                            long long xl, long long xr,
                                            long long& xresult) {
  int p = a[rnd.rand(0, a.size())];
  long long xp = _f(p, v, xl, xr);
  xresult = xp;
  vector<int> vL, vM, vR;
  for (auto idx : a) {
    if (idx == p) {
      vM.push_back(idx);
    } else {
      long long y = query(idx, xp);
      if (y < v) {
        vR.push_back(idx);
      } else if (y > v) {
        vL.push_back(idx);
      } else {
        vM.push_back(idx);
      }
    }
  }
  while ((signed)vL.size() < k && !vM.empty()) {
    vL.push_back(vM.back());
    vM.pop_back();
  }
  while (!vM.empty()) {
    vR.push_back(vM.back());
    vM.pop_back();
  }
  if ((signed)vL.size() == k) {
    return make_pair(vL, vR);
  }
  if ((signed)vL.size() < k) {
    pair<vector<int>, vector<int> > par =
        kth_element(vR, k - vL.size(), v, xp, xr, xresult);
    return make_pair(vL + par.first, par.second);
  }
  pair<vector<int>, vector<int> > par = kth_element(vL, k, v, xl, xp, xresult);
  return make_pair(par.first, vR + par.second);
}
void dividing(vector<int> a, int l, long long xl, long long xr) {
  if ((signed)a.size() == 1) {
    ans[a.back()] = make_pair(xl, xr);
    return;
  }
  int hn = ((signed)a.size()) >> 1;
  long long xm;
  pair<vector<int>, vector<int> > par =
      kth_element(a, hn, (l + hn) * m, xl, xr, xm);
  dividing(par.first, l, xl, xm);
  dividing(par.second, l + hn, xm, xr);
}
int main() {
  scanf("%d%lld", &n, &L);
  m = L / n;
  vector<int> idx(n);
  for (int i = 0; i < n; i++) idx[i] = i + 1;
  dividing(idx, 0, 0, 1e18);
  puts("!");
  for (int i = 1; i <= n; i++) {
    printf("%lld %lld\n", ans[i].first, ans[i].second);
  }
  fflush(stdout);
  return 0;
}