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 N = 1005;
inline long long query(int id, long long x) {
  if (!x) return 0;
  printf("? %d %lld\n", id, x);
  fflush(stdout);
  long long ret;
  scanf("%lld", &ret);
  return ret;
}
inline long long findx(int id, long long L, long long R, long long v) {
  long long mid;
  while (L < R) {
    mid = (L + R) >> 1;
    long long val = query(id, mid);
    if (val < v)
      L = mid + 1;
    else if (v < val)
      R = mid - 1;
    else
      return mid;
  }
  return L;
}
long long tim, vis[N];
long long getmid(vector<int> &f, vector<int> &g, long long L, long long R,
                 long long v, int cnt) {
  int id = f[rand() % f.size()];
  long long x = findx(id, L, R, v), val;
  vector<int> l, r, p;
  for (int i : g) {
    if (i == id) continue;
    val = query(i, x);
    if (val < v)
      l.push_back(i);
    else if (val > v)
      r.push_back(i);
    else
      p.push_back(i);
  }
  while (l.size() < cnt && !p.empty()) l.push_back(p.back()), p.pop_back();
  while (p.size()) r.push_back(p.back()), p.pop_back();
  if (l.size() <= cnt && r.size() < g.size() - cnt) {
    f.clear();
    return x;
  }
  if (l.size() > cnt) {
    ++tim;
    for (int i = 0; i < l.size(); ++i) vis[l[i]] = tim;
    for (int i = 0; i < f.size(); i++)
      if (vis[f[i]] != tim) swap(f[i], f.back()), f.pop_back(), i--;
    return getmid(f, g, L, R, v, cnt);
  } else {
    ++tim;
    for (int i = 0; i < r.size(); ++i) vis[r[i]] = tim;
    for (int i = 0; i < f.size(); i++)
      if (vis[f[i]] != tim) swap(f[i], f.back()), f.pop_back(), i--;
    return getmid(f, g, L, R, v, cnt);
  }
}
struct Interval {
  long long l, r;
} ans[N];
void work(vector<int> &f, long long L, long long R, long long xl,
          long long xr) {
  if (!f.size()) return;
  if (f.size() == 1) {
    ans[f[0]].l = xl;
    ans[f[0]].r = xr;
    return;
  }
  long long m = f.size();
  long long d = (R - L) / m;
  vector<int> g = f;
  long long x = getmid(g, f, xl, xr, L + d * (m / 2), m >> 1);
  long long val;
  vector<int> l, r, p;
  for (int i : f) {
    val = query(i, x);
    if (val > L + d * (m / 2))
      l.push_back(i);
    else if (val < L + d * (m / 2))
      r.push_back(i);
    else
      p.push_back(i);
  }
  while (l.size() < (m >> 1) && p.size()) l.push_back(p.back()), p.pop_back();
  while (!p.empty()) r.push_back(p.back()), p.pop_back();
  work(l, L, L + d * (m / 2), xl, x);
  work(r, L + d * (m / 2), R, x, xr);
}
vector<int> f;
int n;
long long L;
int main() {
  scanf("%d %lld", &n, &L);
  for (int i = 1; i <= n; ++i) f.push_back(i);
  work(f, 0, L, 0, 1e18);
  putchar('!');
  putchar('\n');
  for (int i = 1; i <= n; ++i) printf("%lld %lld\n", ans[i].l, ans[i].r);
  return 0;
}