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>
const int64_t BILLION = 1000000000LL;
const int64_t MAXIN = BILLION * BILLION;
int64_t N, L;
std::map<std::pair<int, int64_t>, int> memo;
int64_t query(int i, int64_t x) {
  if (x == 0) return 0;
  if (x == MAXIN) return L;
  printf("? %d %I64d\n", i + 1, x);
  fflush(stdout);
  int64_t res;
  scanf("%I64d", &res);
  return res;
}
std::pair<int64_t, int64_t> ans[1005];
int64_t locate(int i, int64_t ll, int64_t rr, int64_t thres) {
  int64_t low = ll, high = rr;
  while (high - low > 1) {
    int64_t mid = (low + high) / 2;
    if (query(i, mid) >= thres) {
      high = mid;
    } else {
      low = mid;
    }
  }
  return high;
}
void solve(std::vector<int> is, int64_t ll, int64_t rr, int64_t base) {
  if (is.size() == 1) {
    ans[is[0]] = {ll, rr};
    return;
  }
  int64_t crit = L / N * (is.size() / 2) + base;
  int64_t k = is.size() / 2;
  std::vector<int> sub(is);
  std::vector<int> left, right;
  int64_t pivot_value;
  while (k > 0) {
    int pivot_index = sub[rand() % sub.size()];
    pivot_value = locate(pivot_index, ll, rr, crit);
    std::vector<int> small, big;
    for (int i : sub) {
      if (i >= pivot_index && (query(i, pivot_value - 1) < crit) ||
          i < pivot_index && (query(i, pivot_value) < crit)) {
        big.push_back(i);
      } else {
        small.push_back(i);
      }
    }
    if (k >= small.size()) {
      k -= small.size();
      left.insert(left.end(), small.begin(), small.end());
      sub = big;
    } else {
      right.insert(right.end(), big.begin(), big.end());
      sub = small;
    }
  }
  right.insert(right.end(), sub.begin(), sub.end());
  assert(left.size() <= is.size() / 2);
  assert(right.size() <= is.size() - is.size() / 2);
  solve(left, ll, pivot_value, base);
  solve(right, pivot_value, rr, crit);
}
int main() {
  scanf("%I64d %I64d", &N, &L);
  std::vector<int> is;
  for (int i = 0; i < N; i++) {
    is.push_back(i);
  }
  solve(is, 0, MAXIN, 0);
  printf("!\n");
  for (int i = 0; i < N; i++) {
    printf("%I64d %I64d\n", ans[i].first, ans[i].second);
  }
  return 0;
}