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;
inline long long qry(int id, long long x) {
  cout << "? " << id << " " << x << endl;
  long long ans;
  cin >> ans;
  return ans;
}
long long get(int x, int d, long long l, long long r) {
  while (l < r) {
    long long mid = (l + r) >> 1;
    if (qry(x, mid) >= d * L)
      r = mid;
    else
      l = mid + 1;
  }
  return l;
}
const int N = 2010;
int a[N];
long long pos[N], val;
void getKth(int l, int r, int k, int mid, long long nl, long long nr) {
  static int q1[N], q2[N], q3[N];
  int x = l + rand() % (r - l + 1);
  long long v = get(a[x], mid, nl, nr);
  int s1 = 0, s2 = 0, s3 = 0;
  q2[++s2] = a[x];
  for (int i = l; i <= r; i++)
    if (i != x) {
      long long t = qry(a[i], v);
      if (t > mid * L)
        q1[++s1] = a[i];
      else if (t < mid * L)
        q3[++s3] = a[i];
      else
        q2[++s2] = a[i];
    }
  for (int i = 1; i <= s1; i++) a[l + i - 1] = q1[i];
  for (int i = 1; i <= s2; i++) a[l + s1 + i - 1] = q2[i];
  for (int i = 1; i <= s3; i++) a[l + s1 + s2 + i - 1] = q3[i];
  if (s1 >= k)
    getKth(l, l + s1 - 1, k, mid, nl, v);
  else if (s1 + s2 < k)
    getKth(l + s1 + s2, r, k - s1 - s2, mid, v, nr);
  else
    val = v;
}
inline void solve(int l, int r, long long nl, long long nr) {
  if (l == r) {
    pos[l] = get(a[l], l, nl, nr);
    return;
  }
  int mid = (l + r) >> 1;
  getKth(l, r, mid - l + 1, mid, nl, nr);
  long long t = val;
  solve(l, mid, nl, t);
  solve(mid + 1, r, t, nr);
}
pair<long long, long long> ans[N];
int main() {
  srand(time(0));
  cin >> n >> L;
  L /= n;
  for (int i = 1; i <= n; i++) a[i] = i;
  solve(1, n, 1, 1e18);
  for (int i = 1; i <= n; i++)
    ans[a[i]] = pair<long long, long long>(pos[i - 1], pos[i]);
  puts("!");
  for (int i = 1; i <= n; i++)
    printf("%lld %lld\n", ans[i].first, ans[i].second);
  fflush(stdout);
  return 0;
}