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 n;
long long L;
long long func(int i, long long x) {
  printf("? %d %I64d\n", i, x);
  fflush(stdout);
  long long ans;
  scanf("%I64d", &ans);
  return ans;
}
long long _find(int i, long long val, long long lo, long long hi) {
  while (hi - lo > 1) {
    long long me = lo + hi >> 1;
    long long cur = func(i, me);
    if (cur == val)
      return me;
    else if (cur < val)
      lo = me;
    else
      hi = me;
  }
  return lo;
}
int id[1010], jd[1010];
long long le[1010], ri[1010];
int Rand(int up) { return (rand() << 15 | rand()) % up; }
long long _find(long long lo, long long hi, int le, int ri, int me,
                long long tar) {
  if (le == ri) return _find(id[le], tar, lo, hi);
  int cur = Rand(ri - le + 1) + le;
  assert(le <= cur && cur <= ri);
  int tle = le, tri = ri;
  long long pos = _find(id[cur], tar, lo, hi);
  vector<int> tmp;
  tmp.push_back(id[cur]);
  for (int i = le; i <= ri; i++) {
    if (i == cur) continue;
    long long now = func(id[i], pos);
    if (now > tar)
      jd[tle++] = id[i];
    else if (now < tar)
      jd[tri--] = id[i];
    else
      tmp.push_back(id[i]);
  }
  for (int i = 0; i < tmp.size(); i++) jd[tle + i] = tmp[i];
  for (int i = le; i <= ri; i++) id[i] = jd[i];
  if (tle > me)
    return _find(lo, pos + 1, le, tle - 1, me, tar);
  else if (tri < me)
    return _find(pos, hi, tle + 1, ri, me, tar);
  else
    return pos;
}
void solve(int l, int r, long long lo, long long hi) {
  if (l == r) {
    le[id[l]] = lo;
    ri[id[l]] = hi - 1;
    return;
  }
  int me = l + r >> 1;
  long long pos = _find(lo, hi, l, r, me, L * me);
  solve(l, me, lo, pos + 1);
  solve(me + 1, r, pos, hi);
}
int main() {
  cin >> n >> L;
  L /= n;
  long long up = 1e18;
  for (int i = 1; i <= n; i++) id[i] = i;
  solve(1, n, 0, up + 1);
  puts("!");
  for (int i = 1; i <= n; i++) printf("%I64d %I64d\n", le[i], ri[i]);
  return 0;
}