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 inf = (int)1.01e9;
const long long infll = (long long)1.01e18;
const long double eps = 1e-9;
const long double pi = acos((long double)-1);
mt19937 mrand(chrono::steady_clock::now().time_since_epoch().count());
int rnd(int x) { return mrand() % x; }
void precalc() {}
const int maxn = 1005;
int n;
long long s;
bool read() {
  if (scanf("%d%lld", &n, &s) < 2) {
    return false;
  }
  return true;
}
const long long X = (long long)1e18;
int p[maxn];
long long vals[maxn];
long long ls[maxn], rs[maxn];
long long ask(int i, long long x) {
  printf("? %d %lld\n", i + 1, x);
  fflush(stdout);
  long long res;
  scanf("%lld", &res);
  return res;
}
long long get(int i, int k) {
  long long y = s / n * k;
  long long l = 0, r = X;
  while (l < r - 1) {
    long long m = (l + r) / 2;
    if (ask(p[i], m) < y) {
      l = m;
    } else {
      r = m;
    }
  }
  return r;
}
long long getKth(int l, int m, int r) {
  if (l == r - 1) {
    return get(l, l);
  }
  int k = l + rnd(r - l);
  long long x = get(k, m);
  long long y = s / n * m;
  int m0 = l, m1 = l;
  for (int i = l; i < r; i++) {
    vals[i] = ask(p[i], x);
    if (vals[i] > y) {
      swap(p[i], p[m0]);
      swap(vals[i], vals[m0]);
      if (m1 == m0) {
        m1++;
      }
      m0++;
    }
    if (vals[i] == y) {
      swap(p[i], p[m1]);
      swap(vals[i], vals[m1]);
      m1++;
    }
  }
  if (m0 <= m && m < m1) {
    return x;
  }
  if (m < m0) {
    return getKth(l, m, m0);
  } else {
    return getKth(m1, m, r);
  }
}
void rec(int l, int r, long long lx, long long rx) {
  if (l == r - 1) {
    ls[p[l]] = lx;
    rs[p[l]] = rx;
    return;
  }
  int m = (l + r) / 2;
  long long mx = getKth(l, m, r);
  rec(l, m, lx, mx);
  rec(m, r, mx, rx);
}
void solve() {
  for (int i = 0; i < n; i++) {
    p[i] = i;
  }
  rec(0, n, 0, X);
  printf("!\n");
  for (int i = 0; i < n; i++) {
    printf("%lld %lld\n", ls[i], rs[i]);
  }
  fflush(stdout);
}
int main() {
  precalc();
  read();
  solve();
  return 0;
}