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 ul = std::uint32_t;
using li = std::int32_t;
using ll = std::int64_t;
using ull = std::uint64_t;
using vul = std::vector<ul>;
using vvul = std::vector<vul>;
using pulul = std::pair<ul, ul>;
ul n;
ull L;
std::mt19937 rands;
class func {
 public:
  ul id = 0;
  ull x = 0;
  ull x_ = 0;
};
func data[1001];
void deal(ul nl, ul nr, ul tar, ull lx, ull rx) {
  if (nr + 1 == nl) {
    return;
  }
  ul key = nl + rands() % (nr - nl + 1);
  ull l = lx, r = rx;
  while (l + 1 != r) {
    ull mid = l + r >> 1;
    std::printf("? %u %llu\n", data[key].id, mid);
    std::fflush(stdout);
    ull val;
    std::scanf("%llu", &val);
    if (val <= tar * (L / n)) {
      l = mid;
    } else {
      r = mid;
    }
  }
  data[key].x = l;
  ull xf = tar * (L / n);
  std::swap(data[key], data[nl]);
  ul i = nl + 1, j = nr;
  for (; i <= j;) {
    std::printf("? %u %llu\n", data[i].id, data[nl].x);
    std::fflush(stdout);
    ull val;
    std::scanf("%llu", &val);
    if (val < xf) {
      std::swap(data[i], data[j]);
      --j;
    } else if (val > xf) {
      ++i;
    } else {
      if (data[nl].x == 1e18) {
        if (data[i].id > data[nl].id) {
          std::swap(data[i], data[j]);
          --j;
        } else {
          ++i;
        }
      } else {
        std::printf("? %u %llu\n", data[i].id, data[nl].x + 1);
        std::fflush(stdout);
        ull val;
        std::scanf("%llu", &val);
        if (val == xf || data[i].id > data[nl].id) {
          std::swap(data[i], data[j]);
          --j;
        } else {
          ++i;
        }
      }
    }
  }
  std::swap(data[nl], data[j]);
  if (j < tar) {
    deal(j + 1, nr, tar, data[j].x, rx);
  } else if (j > tar) {
    deal(nl, j - 1, tar, lx, data[j].x + 1);
  }
}
void deal(ul nl, ul nr, ull lx, ull rx) {
  if (nr + 1 == nl) {
    return;
  }
  ul nmid = nr + nl >> 1;
  deal(nl, nr, nmid, lx, rx);
  if (nmid != nl) {
    deal(nl, nmid, lx, data[nmid].x + 1);
  }
  deal(nmid + 1, nr, data[nmid].x + 1, rx);
}
int main() {
  std::scanf("%u%llu", &n, &L);
  rands.seed(std::time(0));
  for (ul i = 1; i <= n; ++i) {
    data[i].id = i;
  }
  deal(1, n, 0, ull(1e18) + 1);
  std::printf("!\n");
  std::fflush(stdout);
  for (ul i = 1; i <= n; ++i) {
    data[i].x_ = data[i - 1].x;
  }
  std::sort(data + 1, data + n + 1,
            [](const func& a, const func& b) { return a.id < b.id; });
  for (ul i = 1; i <= n; ++i) {
    std::printf("%llu %llu\n", data[i].x_, data[i].x);
    std::fflush(stdout);
  }
  return 0;
}