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;
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
const int MAXN = 1000;
const long long XMIN = 0LL;
const long long XMAX = 1000000000000000000LL;
mt19937 rnd((int)std::chrono::steady_clock::now().time_since_epoch().count());
int n;
long long ymax;
long long lans[MAXN], rans[MAXN];
long long eval(int idx, long long x) {
  printf("? %d %lld\n", idx + 1, x);
  fflush(stdout);
  long long y;
  scanf("%lld", &y);
  return y;
}
long long findx(int idx, long long y, long long lx, long long hx) {
  --lx;
  while (lx + 1 < hx) {
    long long mx = lx + (hx - lx) / 2;
    long long cy = eval(idx, mx);
    if (cy < y)
      lx = mx;
    else
      hx = mx;
  }
  return hx;
}
void rec(const vector<int> &rem, long long lx, long long hx, long long ly,
         long long hy) {
  if (((int)(rem).size()) == 1) {
    lans[rem[0]] = lx, rans[rem[0]] = hx;
    return;
  }
  vector<int> cand = rem;
  int n1 = ((int)(rem).size()) / 2, n2 = ((int)(rem).size()) - n1;
  assert((hy - ly) % (n1 + n2) == 0);
  long long mx = -1, my = ly + (hy - ly) / (n1 + n2) * n1;
  vector<int> lft, rgt;
  while (true) {
    int cur = cand[rnd() % ((int)(cand).size())];
    long long cx = findx(cur, my, lx, hx);
    vector<int> candbefore, candeq, candafter;
    candeq.push_back(cur);
    for (int i = (0); i < (((int)(cand).size())); ++i) {
      int idx = cand[i];
      if (idx == cur) continue;
      long long cy = eval(idx, cx);
      if (cy < my)
        candafter.push_back(idx);
      else if (cy > my)
        candbefore.push_back(idx);
      else
        candeq.push_back(idx);
    }
    if (((int)(lft).size()) + ((int)(candbefore).size()) > n1) {
      for (int i = (0); i < (((int)(candafter).size())); ++i)
        rgt.push_back(candafter[i]);
      for (int i = (0); i < (((int)(candeq).size())); ++i)
        rgt.push_back(candeq[i]);
      cand = candbefore;
      continue;
    }
    if (((int)(rgt).size()) + ((int)(candafter).size()) > n2) {
      for (int i = (0); i < (((int)(candbefore).size())); ++i)
        lft.push_back(candbefore[i]);
      for (int i = (0); i < (((int)(candeq).size())); ++i)
        lft.push_back(candeq[i]);
      cand = candafter;
      continue;
    }
    mx = cx;
    for (int i = (0); i < (((int)(candbefore).size())); ++i)
      lft.push_back(candbefore[i]);
    for (int i = (0); i < (((int)(candafter).size())); ++i)
      rgt.push_back(candafter[i]);
    for (int i = (0); i < (((int)(candeq).size())); ++i)
      if (((int)(lft).size()) < n1)
        lft.push_back(candeq[i]);
      else
        rgt.push_back(candeq[i]);
    assert(((int)(lft).size()) == n1 && ((int)(rgt).size()) == n2);
    break;
  }
  rec(lft, lx, mx, ly, my);
  rec(rgt, mx, hx, my, hy);
}
void solve() {
  vector<int> all(n);
  for (int i = (0); i < (n); ++i) all[i] = i;
  rec(all, XMIN, XMAX, 0, ymax);
}
void run() {
  scanf("%d%lld", &n, &ymax);
  solve();
  printf("!\n");
  for (int i = (0); i < (n); ++i) printf("%lld %lld\n", lans[i], rans[i]);
  fflush(stdout);
}
int main() {
  run();
  return 0;
}