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;
mt19937 rnd(20030102);
long long query(int id, long long x) {
  printf("? %d %lld\n", id, x);
  fflush(stdout);
  long long ret;
  scanf("%lld", &ret);
  return ret;
}
int n;
long long lres[1010], rres[1010];
long long L;
long long get(int id, long long vl) {
  long long L = 0, R = 1000000000000000000, pos = -1;
  while (L <= R) {
    long long mid = (L + R) >> 1;
    if (query(id, mid) >= vl)
      pos = mid, R = mid - 1;
    else
      L = mid + 1;
  }
  return pos;
}
pair<int, long long> getkth(vector<int> dat, int K, long long mpos) {
  int p = rnd() % dat.size();
  long long pos = get(dat[p], mpos);
  vector<int> ls, rs, ms;
  for (int i = 0; i < dat.size(); i++) {
    if (i == p) continue;
    long long c = query(dat[i], pos);
    if (c == mpos)
      ms.push_back(dat[i]);
    else if (c > mpos)
      ls.push_back(dat[i]);
    else
      rs.push_back(dat[i]);
  }
  while (ls.size() < K - 1 && !ms.empty()) {
    ls.push_back(ms.back());
    ms.pop_back();
  }
  while (!ms.empty()) {
    rs.push_back(ms.back());
    ms.pop_back();
  }
  if (ls.size() == K - 1)
    return make_pair(dat[p], pos);
  else if (ls.size() < K - 1)
    return getkth(rs, K - ls.size() - 1, mpos);
  else
    return getkth(ls, K, mpos);
}
void solve(vector<int> s, long long l, long long r, long long lv) {
  if (s.size() == 1) {
    lres[s.front()] = l;
    rres[s.front()] = r;
    return;
  }
  long long midv = lv + (L / n) * (s.size() / 2);
  pair<int, long long> mm = getkth(s, s.size() / 2, midv);
  vector<int> ls, rs, ms;
  for (int i = 0; i < s.size(); i++) {
    long long c = query(s[i], mm.second);
    if (c > midv)
      ls.push_back(s[i]);
    else if (c == midv)
      ms.push_back(s[i]);
    else
      rs.push_back(s[i]);
  }
  while (ls.size() < s.size() / 2) {
    ls.push_back(ms.back());
    ms.pop_back();
  }
  while (!ms.empty()) {
    rs.push_back(ms.back());
    ms.pop_back();
  }
  solve(ls, l, mm.second, lv);
  solve(rs, mm.second, r, midv);
}
int main() {
  scanf("%d%lld", &n, &L);
  vector<int> res(n);
  iota(res.begin(), res.end(), 1);
  solve(res, 0, 1000000000000000000, 0);
  puts("!");
  for (int i = 1; i <= n; i++) printf("%lld %lld\n", lres[i], rres[i]);
  return 0;
}