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 = 1e3 + 5;
const long long Range = 1e18;
template <typename T>
void chkmax(T &x, T y) {
  x = max(x, y);
}
template <typename T>
void chkmin(T &x, T y) {
  x = min(x, y);
}
template <typename T>
void read(T &x) {
  x = 0;
  int f = 1;
  char c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') f = -f;
  for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
  x *= f;
}
int n, cnt, a[MAXN];
long long L, b[MAXN], c[MAXN];
pair<long long, long long> ans[MAXN];
long long locate(int i, long long x, long long l, long long r) {
  while (l < r) {
    long long mid = (l + r) / 2, res;
    cout << '?' << ' ' << i << ' ' << mid << endl;
    read(res);
    if (res == x) return mid;
    if (res > x)
      r = mid - 1;
    else
      l = mid + 1;
  }
  return l;
}
void findkth(int l, int r, int k, long long x, long long pl, long long pr) {
  random_shuffle(a + l, a + r + 1);
  b[a[l]] = locate(a[l], x, pl, pr);
  long long pos = b[a[l]];
  for (int i = l; i <= r; i++) {
    cout << '?' << ' ' << a[i] << ' ' << pos << endl;
    read(c[a[i]]);
    if (c[a[i]] == x) b[a[i]] = pos;
  }
  sort(a + l, a + r + 1, [&](int x, int y) { return c[x] > c[y]; });
  if (c[a[k]] == x) return;
  if (c[a[k]] < x) {
    while (c[a[l]] >= x) l++;
    findkth(l, r, k, x, pl, pr);
  } else {
    while (c[a[r]] <= x) r--;
    findkth(l, r, k, x, pl, pr);
  }
}
void solve(int l, int r, long long vl, long long vr, long long pl,
           long long pr) {
  if (l == r) {
    ans[a[l]] = make_pair(pl, pr);
    return;
  }
  assert((vr - vl) % (r - l + 1) == 0);
  long long d = (vr - vl) / (r - l + 1), k = (r - l + 1) / 2;
  findkth(l, r, l + k - 1, vl + k * d, pl, pr);
  long long mid = b[a[l + k - 1]];
  solve(l, l + k - 1, vl, vl + k * d, pl, mid);
  solve(l + k, r, vl + k * d, vr, mid, pr);
}
int main() {
  read(n), read(L);
  for (int i = 1; i <= n; i++) a[i] = i;
  solve(1, n, 0, L, 0, Range);
  cout << '!' << endl;
  for (int i = 1; i <= n; i++)
    cout << ans[i].first << ' ' << ans[i].second << endl;
  return 0;
}