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;
const int MAXN = 1005;
struct result {
  long long l, r;
} ans[MAXN];
mt19937 gen(time(NULL));
long long L;
int id[MAXN], tmp[MAXN];
long long query(int f, long long p) {
  printf("? %d %lld\n", f, p);
  fflush(stdout);
  long long v;
  scanf("%lld", &v);
  return v;
}
long long get_pos(int x, long long l, long long r, long long v) {
  while (l <= r) {
    long long m = l + r >> 1;
    long long val = query(x, m);
    if (val >= v)
      r = m - 1;
    else
      l = m + 1;
  }
  return l;
}
void solve(int fl, int fr, long long wl, long long wr) {
  if (fl == fr) {
    ans[id[fl]] = {wl, wr};
    return;
  }
  int l = fl, r = fr, m = (fl + fr) / 2;
  long long V = m * L, p;
  for (;;) {
    int x = uniform_int_distribution<int>(l, r)(gen);
    p = get_pos(id[x], wl, wr, V);
    int h1 = l, h2 = r;
    vector<int> eq;
    for (int i = l; i <= r; i++)
      if (i != x) {
        long long val = query(id[i], p);
        if (val < V)
          tmp[h2--] = id[i];
        else if (val > V)
          tmp[h1++] = id[i];
        else
          eq.push_back(id[i]);
      }
    eq.push_back(id[x]);
    int th = h1;
    for (int &x : eq) tmp[th++] = x;
    for (int i = l; i <= r; i++) id[i] = tmp[i];
    if (m >= h1 && m <= h2) break;
    if (m < h1)
      r = h1 - 1;
    else
      l = h2 + 1;
  }
  solve(fl, m, wl, p);
  solve(m + 1, fr, p, wr);
}
int main() {
  int n;
  scanf("%d%lld", &n, &L);
  L /= n;
  for (int i = 1; i <= n; i++) id[i] = i;
  solve(1, n, 0, 1000000000000000000);
  puts("!");
  for (int i = 1; i <= n; i++) printf("%lld %lld\n", ans[i].l, ans[i].r);
  return 0;
}