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;
pair<long long, long long> h[1010];
long long L;
int a[1010], a1[1010], a2[1010], n;
bool vis[1010];
inline long long rd() {
  long long x = 0;
  char ch = getchar();
  for (; ch < '0' || ch > '9'; ch = getchar())
    ;
  for (; ch >= '0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - '0';
  return x;
}
inline long long ask(int x, long long y) {
  printf("? %d %lld\n", x, y);
  fflush(stdout);
  return rd();
}
inline long long calc(int x, long long y, long long ql, long long qr) {
  long long l = ql, r = qr;
  while (l < r) {
    long long mid = ((l + r) >> 1) + 1;
    if (ask(x, mid) <= y)
      l = mid;
    else
      r = mid - 1;
  }
  return l;
}
long long hhh;
inline vector<int> select(vector<int> v, long long hh, int k, long long ql,
                          long long qr) {
  long long mid = calc(v[rand() % v.size()], hh, ql, qr);
  hhh = mid;
  vector<int> v1, v2, v3;
  for (int t : v) {
    long long p = ask(t, mid);
    if (p < hh)
      v3.push_back(t);
    else if (p > hh)
      v1.push_back(t);
    else {
      long long q = ask(t, mid + 1);
      if (q == hh)
        v3.push_back(t);
      else
        v2.push_back(t);
    }
  }
  if (v1.size() >= k)
    return select(v1, hh, k, ql, mid);
  else if (v1.size() + v2.size() < k) {
    vector<int> res;
    for (int t : v1) res.push_back(t);
    for (int t : v2) res.push_back(t);
    vector<int> res1 = select(v3, hh, k - v1.size() - v2.size(), mid, qr);
    for (int t : res1) res.push_back(t);
    return res;
  } else {
    vector<int> res;
    for (int t : v1) res.push_back(t);
    for (int i = 0; i < k - v1.size(); i++) res.push_back(v2[i]);
    return res;
  }
}
inline void solve(int l, int r, long long ql, long long qr) {
  if (l == r) {
    h[a[l]].first = ql;
    h[a[l]].second = qr;
    return;
  }
  vector<int> v;
  for (int i = l; i <= r; i++) v.push_back(a[i]);
  int mid = (l + r) >> 1;
  long long hh = L / n * mid;
  vector<int> g = select(v, hh, mid - l + 1, ql, qr);
  for (int i = l; i <= r; i++) vis[a[i]] = false;
  for (int t : g) vis[t] = true;
  int t1 = 0, t2 = 0;
  for (int i = l; i <= r; i++)
    if (vis[a[i]])
      a1[++t1] = a[i];
    else
      a2[++t2] = a[i];
  for (int i = 1; i <= t1; i++) a[l + i - 1] = a1[i];
  for (int i = 1; i <= t2; i++) a[l + t1 + i - 1] = a2[i];
  long long now = hhh;
  solve(l, mid, ql, now);
  solve(mid + 1, r, now, qr);
}
int main() {
  srand(19260816);
  n = rd();
  L = rd();
  for (int i = 1; i <= n; i++) a[i] = i;
  solve(1, n, 0, 1000000000000000000LL);
  puts("!");
  for (int i = 1; i <= n; i++) printf("%lld %lld\n", h[i].first, h[i].second);
  fflush(stdout);
  return 0;
}