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;
int n;
long long L, al[1005], ar[1005];
long long getv(int u, long long v) {
  printf("? %d %lld\n", u, v);
  fflush(stdout);
  long long r;
  scanf("%lld", &r);
  return r;
}
long long ser(int u, long long v, long long l, long long r) {
  long long m;
  for (; l < r;) {
    m = l + r >> 1;
    long long t = getv(u, m);
    if (t < v)
      l = m + 1;
    else if (t > v)
      r = m - 1;
    else
      return m;
  }
  return l;
}
int vis[1005], tim;
long long llc;
void qry(vector<int>& g, vector<int>& p, long long ql, long long qr,
         long long aim, int ls) {
  int t = g[rand() % g.size()];
  long long x = ser(t, aim, ql, qr);
  vector<int> l, r, z;
  for (int i = (0), LIM = (p.size() - 1); i <= LIM; i++)
    if (p[i] != t) {
      long long y = getv(p[i], x);
      if (y == aim) z.push_back(p[i]);
      if (y < aim) l.push_back(p[i]);
      if (y > aim) r.push_back(p[i]);
    }
  for (; l.size() < ls && !z.empty();) l.push_back(z.back()), z.pop_back();
  for (; !z.empty();) r.push_back(z.back()), z.pop_back();
  if (l.size() <= ls && r.size() < p.size() - ls) {
    g.clear();
    g.push_back(t);
    llc = x;
    return;
  }
  if (l.size() > ls) {
    ++tim;
    for (int i = (0), LIM = (l.size() - 1); i <= LIM; i++) vis[l[i]] = tim;
    for (int i = 0; i < g.size(); i++)
      if (vis[g[i]] != tim) swap(g[i], g.back()), g.pop_back(), i--;
    qry(g, p, ql, qr, aim, ls);
  } else {
    ++tim;
    for (int i = (0), LIM = (r.size() - 1); i <= LIM; i++) vis[r[i]] = tim;
    for (int i = 0; i < g.size(); i++)
      if (vis[g[i]] != tim) swap(g[i], g.back()), g.pop_back(), i--;
    qry(g, p, ql, qr, aim, ls);
  }
}
void Solve(vector<int>& p, long long ql, long long qr, long long pl,
           long long pr) {
  if (!p.size()) return;
  if (p.size() == 1) {
    al[p[0]] = ql, ar[p[0]] = qr;
    return;
  }
  vector<int> g = p;
  int m = p.size() / 2;
  qry(g, p, ql, qr, pl + (pr - pl) / p.size() * m, m);
  vector<int> l, r, z;
  for (int i = (0), LIM = (p.size() - 1); i <= LIM; i++) {
    long long y = getv(p[i], llc);
    if (y == pl + (pr - pl) / p.size() * m) z.push_back(p[i]);
    if (y > pl + (pr - pl) / p.size() * m) l.push_back(p[i]);
    if (y < pl + (pr - pl) / p.size() * m) r.push_back(p[i]);
  }
  for (; l.size() < m;) l.push_back(z.back()), z.pop_back();
  for (; !z.empty();) r.push_back(z.back()), z.pop_back();
  long long tmp = llc;
  Solve(l, ql, tmp, pl, pl + (pr - pl) / p.size() * m);
  Solve(r, tmp, qr, pl + (pr - pl) / p.size() * m, pr);
}
int main() {
  scanf("%d%lld", &n, &L);
  vector<int> p;
  for (int i = (1), LIM = (n); i <= LIM; i++) p.push_back(i);
  Solve(p, 0, 1e18, 0, L);
  puts("!");
  for (int i = (1), LIM = (n); i <= LIM; i++)
    printf("%lld %lld\n", al[i], ar[i]);
}