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;
inline char gc() { return getchar(); }
template <class T>
int read(T &ans) {
  ans = 0;
  char ch = gc();
  T f = 1;
  while (!isdigit(ch)) {
    if (ch == EOF) return -1;
    if (ch == '-') f = -1;
    ch = gc();
  }
  while (isdigit(ch)) ans = ans * 10 + ch - '0', ch = gc();
  ans *= f;
  return 1;
}
template <class T1, class T2>
int read(T1 &a, T2 &b) {
  return read(a) != EOF && read(b) != EOF ? 2 : EOF;
}
template <class T1, class T2, class T3>
int read(T1 &a, T2 &b, T3 &c) {
  return read(a, b) != EOF && read(c) != EOF ? 3 : EOF;
}
const int Maxn = 1100;
const int inf = 0x3f3f3f3f;
int a[Maxn], b[Maxn], c[Maxn], tims;
long long n, L, v;
pair<long long, long long> ans[Maxn];
long long query(long long a, long long b) {
  printf("? %I64d %I64d\n", a, b);
  fflush(stdout);
  long long ans;
  read(ans);
  return ans;
}
long long find(int x, long long k, long long l, long long r) {
  long long mid = l + r >> 1;
  while (1) {
    long long y = query(x, mid);
    if (y == k) return mid;
    if (y < k)
      l = mid + 1;
    else
      r = mid - 1;
    mid = l + r >> 1;
  }
}
void kth(int l, int r, int k, long long ql, long long qr) {
  if (l == r) return;
  int x = rand() % (r - l + 1) + l;
  long long y = find(a[x], (L / n) * k, ql, qr);
  int mid = k - l + 1;
  int l1 = l - 1, l2 = r + 1, l3 = 0;
  c[++l3] = a[x];
  for (int i = l; i <= r; i++) {
    if (i == x) continue;
    long long sxz = query(a[i], y);
    if (sxz < (L / n) * k) b[--l2] = a[i];
    if (sxz == (L / n) * k) c[++l3] = a[i];
    if (sxz > (L / n) * k) b[++l1] = a[i];
  }
  while (l1 < k && l3) b[++l1] = c[l3--];
  if (l1 == k) {
    v = y;
    while (l3) b[--l2] = c[l3--];
    for (int i = l; i <= r; i++) a[i] = b[i];
    return;
  }
  if (l1 < k) {
    for (int i = l; i <= r; i++) a[i] = b[i];
    kth(l1 + 1, r, k, ql, qr);
    return;
  }
  while (l3) b[--l2] = c[l3--];
  for (int i = l; i <= r; i++) a[i] = b[i];
  kth(l, l1, k, ql, qr);
}
void solve(int l, int r, long long ql, long long qr) {
  if (l == r) {
    ans[a[l]] = make_pair(ql, qr);
    return;
  }
  int mid = l + r >> 1;
  kth(l, r, mid, ql, qr);
  long long x = v;
  solve(l, mid, ql, x);
  solve(mid + 1, r, x, qr);
}
signed main() {
  read(n, L);
  for (int i = 1; i <= n; i++) a[i] = i;
  solve(1, n, 0, 1000000000000000000ll);
  puts("!");
  for (int i = 1; i <= n; i++)
    printf("%I64d %I64d\n", ans[i].first, ans[i].second);
  fflush(stdout);
  return 0;
}