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() {
  static char buf[100000], *p1 = buf, *p2 = buf;
  return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2)
             ? EOF
             : *p1++;
}
inline long long read() {
  long long x = 0;
  char ch = getchar();
  bool positive = 1;
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') positive = 0;
  for (; isdigit(ch); ch = getchar()) x = x * 10 + ch - '0';
  return positive ? x : -x;
}
inline void write(long long a) {
  if (a < 0) {
    a = -a;
    putchar('-');
  }
  if (a >= 10) write(a / 10);
  putchar('0' + a % 10);
}
inline void writeln(long long a) {
  write(a);
  puts("");
}
inline void wri(long long a) {
  write(a);
  putchar(' ');
}
inline unsigned long long rnd() {
  return ((unsigned long long)rand() << 30 ^ rand()) << 4 | rand() % 4;
}
const long long inf = 1e18;
const long long N = 1005;
pair<long long, long long> ans[N];
long long L;
long long n;
void hb(vector<long long> &a, vector<long long> &b) {
  for (auto i : b) a.push_back(i);
}
long long ask(long long id, long long pos) {
  putchar('?');
  putchar(' ');
  wri(id);
  writeln(pos);
  fflush(stdout);
  return read();
}
long long get(long long id, long long x, long long l, long long r) {
  while (l < r) {
    long long mid = (l + r) >> 1;
    if (ask(id, mid) >= x)
      r = mid;
    else
      l = mid + 1;
  }
  return l;
}
void solve(long long fl, long long fr, vector<long long> v, long long lll,
           long long rrr) {
  if (v.size() == 1) {
    ans[v[0]] = make_pair(fl, fr);
    return;
  }
  assert(v.size() == rrr - lll + 1);
  long long dq;
  long long dd;
  long long k = (v.size() + 1) >> 1;
  long long mid = (lll + rrr) / 2 * (L / n);
  vector<long long> LL, RR;
  long long gl = fl, gr = fr;
  while (1) {
    vector<long long> jb, jj, md;
    random_shuffle(v.begin(), v.end());
    dq = v[v.size() - 1];
    v.pop_back();
    dd = get(dq, mid, fl, fr);
    md.push_back(dq);
    assert(k && k <= v.size() + 1);
    for (auto i : v) {
      long long t = ask(i, dd);
      if (t == mid)
        md.push_back(i);
      else if (t > mid)
        jb.push_back(i);
      else
        jj.push_back(i);
    }
    if (k > jb.size() && k <= md.size() + jb.size()) {
      hb(LL, jb);
      for (long long i = (long long)(1); i <= (long long)(k - jb.size()); i++) {
        LL.push_back(md.back());
        md.pop_back();
      }
      hb(RR, md);
      hb(RR, jj);
      break;
    } else if (k <= jb.size()) {
      hb(RR, md);
      hb(RR, jj);
      swap(v, jb);
      gr = dd;
    } else {
      hb(LL, md);
      hb(LL, jb);
      k -= md.size() + jb.size();
      swap(v, jj);
      gl = dd + 1;
    }
  }
  solve(fl, dd, LL, lll, (lll + rrr) / 2);
  solve(dd, fr, RR, (lll + rrr) / 2 + 1, rrr);
}
signed main() {
  n = read();
  L = read();
  vector<long long> v;
  for (long long i = (long long)(1); i <= (long long)(n); i++) v.push_back(i);
  solve(0, inf, v, 1, n);
  puts("!");
  for (long long i = (long long)(1); i <= (long long)(n); i++) {
    wri(ans[i].first);
    writeln(ans[i].second);
  }
  fflush(stdout);
}