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;
template <typename T>
void maxtt(T& t1, T t2) {
  t1 = max(t1, t2);
}
template <typename T>
void mintt(T& t1, T t2) {
  t1 = min(t1, t2);
}
bool debug = 0;
int n, m, k;
int dx[4] = {0, 1, 0, -1}, dy[4] = {1, 0, -1, 0};
string direc = "URDL";
const long long MOD2 = (long long)998244353 * (long long)998244353;
long long ln, lk, lm;
void etp(bool f = 0) {
  puts(f ? "YES" : "NO");
  exit(0);
}
void addmod(int& x, int y, int mod = 998244353) {
  assert(y >= 0);
  x += y;
  if (x >= mod) x -= mod;
  assert(x >= 0 && x < mod);
}
void et(int x = -1) {
  printf("%d\n", x);
  exit(0);
}
long long fastPow(long long x, long long y, int mod = 998244353) {
  long long ans = 1;
  while (y > 0) {
    if (y & 1) ans = (x * ans) % mod;
    x = x * x % mod;
    y >>= 1;
  }
  return ans;
}
long long gcd1(long long x, long long y) { return y ? gcd1(y, x % y) : x; }
long long rL;
int a[1035], b[1035];
long long ansl[1035], ansr[1035];
long long rd(int id, long long x) {
  printf("? %d %lld\n", id, x);
  fflush(stdout);
  scanf("%lld", &lk);
  return lk;
}
void cal(int l, int r, long long L, long long R) {
  if (l == r) {
    ansl[a[l]] = L;
    ansr[a[l]] = R;
    return;
  }
  int m1 = (l + r) / 2;
  long long X = rL / n * m1, m2;
  mt19937 ttrand((int)time(0));
  for (int p = l, q = r;;) {
    int i = uniform_int_distribution<int>(p, q)(ttrand);
    long long lb = L, rb = R;
    while (lb < rb) {
      long long mb = (lb + rb) / 2;
      if (rd(a[i], mb) < X)
        lb = mb + 1;
      else
        rb = mb;
    }
    int l2 = p, r2 = q;
    vector<int> e;
    for (int j = p; j <= q; ++j) {
      long long y = rd(a[j], lb);
      if (y > X)
        b[l2++] = a[j];
      else if (y < X)
        b[r2--] = a[j];
      else
        e.push_back(a[j]);
    }
    for (int(j) = 0; (j) < (int)(e.size()); (j)++) b[l2 + j] = e[j];
    for (int z = p; z <= q; z++) a[z] = b[z];
    if (l2 <= m1 && m1 <= r2) {
      m2 = lb;
      break;
    }
    if (l2 < m1)
      p = l2 + 1;
    else
      q = l2 - 1;
  }
  cal(l, m1, L, m2);
  cal(m1 + 1, r, m2, R);
}
void fmain(int tid) {
  scanf("%d%lld", &n, &rL);
  for (int(i) = 1; (i) <= (int)(n); (i)++) a[i] = i;
  cal(1, n, 0, 1e18);
  puts("!");
  for (int(i) = 1; (i) <= (int)(n); (i)++)
    printf("%lld %lld\n", ansl[i], ansr[i]);
}
int main() {
  int t = 1;
  for (int(i) = 1; (i) <= (int)(t); (i)++) {
    fmain(i);
  }
  return 0;
}