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 long long S = (long long)1e18;
long long n, L, p[2005], ansl[2005], ansr[2005], s1[2005], s2[2005], s3[2005];
long long rd() {
  long long x = 0, tp = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') tp = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = (x * 10 + c - '0');
    c = getchar();
  }
  return x * tp;
}
long long query(long long id, long long x) {
  printf("? %lld %lld\n", id, x);
  fflush(stdout);
  long long res;
  scanf("%lld", &res);
  return res;
}
long long getpos(long long id, long long x) {
  long long l = 0, r = S, mid, pos = 0;
  while (l <= r) {
    mid = (l + r) >> 1;
    if (query(id, mid) >= x)
      r = mid - 1;
    else
      l = mid + 1, pos = mid;
  }
  return pos + 1;
}
long long get_kth(long long l, long long r, long long K) {
  long long pos, val, i, W, nvl, t1, t2, t3, top;
  pos = rand() % (r - l + 1) + l;
  W = L / n * K;
  val = getpos(p[pos], W);
  t1 = t2 = t3 = 0;
  for (i = l; i <= r; i++) {
    nvl = query(p[i], val);
    if (nvl > W)
      s1[++t1] = p[i];
    else if (nvl == W)
      s2[++t2] = p[i];
    else
      s3[++t3] = p[i];
  }
  top = l;
  for (i = 1; i <= t1; i++) p[top++] = s1[i];
  for (i = 1; i <= t2; i++) p[top++] = s2[i];
  for (i = 1; i <= t3; i++) p[top++] = s3[i];
  if (t1 > K - l)
    return get_kth(l, l + t1 - 1, K);
  else if (t3 > r - K)
    return get_kth(r - t3 + 1, r, K);
  else
    return val;
}
void solve(long long l, long long r, long long sl, long long sr) {
  if (l == r) {
    ansl[p[l]] = sl, ansr[p[r]] = sr;
    return;
  }
  long long k = (l + r) >> 1, mid;
  mid = get_kth(l, r, k);
  solve(l, k, sl, mid), solve(k + 1, r, mid, sr);
}
int main() {
  scanf("%lld %lld", &n, &L);
  for (int i = 1; i <= n; i++) p[i] = i;
  solve(1, n, 0, S);
  printf("!\n");
  for (int i = 1; i <= n; i++) printf("%lld %lld\n", ansl[i], ansr[i]);
  fflush(stdout);
  return 0;
}