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;
int const N = 1000;
long long const inf = 1000000000000000000;
int n;
long long L;
pair<long long, long long> an[N + 1];
vector<int> o{0};
map<long long, long long> dp[N + 1];
long long ask(int i, long long x) {
  auto it = dp[i].find(x);
  if (it != dp[i].end()) return it->second;
  printf("? %d %lld\n", i, x);
  fflush(stdout);
  long long an;
  scanf("%lld", &an);
  return dp[i][x] = an;
}
long long gt(int i, long long x, long long l, long long r) {
  while (r > l) {
    long long m = l + r >> 1;
    if (ask(i, m) < x)
      l = m + 1;
    else
      r = m;
  }
  return l;
}
void go(vector<int> const &v, long long l = 0, long long r = inf, int nl = 1,
        int nr = n) {
  if (v.size() == 1) {
    an[v[0]].second = r;
    o.push_back(v[0]);
    return;
  }
  int m = nl + nr >> 1;
  long long w = L / n * m, t;
  vector<int> vl, vr, c(v);
  while (!c.empty()) {
    int i = rand() % c.size(), cl = 0, cr = 0;
    t = gt(c[i], w, l, r);
    for (int u : c) {
      long long z = ask(u, t);
      if (z < w)
        ++cl;
      else if (z > w)
        ++cr;
    }
    bool kll = false, klr = false;
    if (cl + vl.size() <= nr - m && cr + vr.size() <= m + 1 - nl) {
      kll = true;
      klr = true;
    } else if (cl + vl.size() > nr - m)
      klr = true;
    else
      kll = true;
    for (int &u : c) {
      long long z = ask(u, t);
      if (z < w && kll)
        vl.push_back(u), u = 0, --cl;
      else if (z > w && klr)
        vr.push_back(u), u = 0, --cr;
      else if (z == w) {
        if (kll && klr) {
          if (cl + vl.size() < nr - m)
            vl.push_back(u);
          else
            vr.push_back(u);
        } else if (kll)
          vl.push_back(u);
        else
          vr.push_back(u);
        u = 0;
      }
    }
    for (int i = 0; i < (int)(c.size()); ++i)
      if (c[i] == 0) {
        swap(c[i], c.back());
        c.pop_back();
        --i;
      }
  }
  go(vr, l, t, nl, m);
  go(vl, t + 1, r, m + 1, nr);
}
int main() {
  srand(time(0));
  scanf("%d%lld", &n, &L);
  vector<int> v(n);
  for (int i = 1; i < (int)(n + 1); ++i) v[i - 1] = i;
  go(v);
  for (int i = 1; i < (int)(n + 1); ++i) an[o[i]].first = an[o[i - 1]].second;
  printf("!\n");
  for (int i = 1; i < (int)(n + 1); ++i)
    printf("%lld %lld\n", an[i].first, an[i].second);
}