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 <class T>
inline void read(T &x) {
  x = 0;
  char c = getchar();
  int f = 1;
  while (!isdigit(c)) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (isdigit(c)) {
    x = x * 10 + c - '0';
    c = getchar();
  }
  x *= f;
}
template <class T>
inline void umin(T &x, T y) {
  x = x < y ? x : y;
}
template <class T>
inline void umax(T &x, T y) {
  x = x > y ? x : y;
}
inline unsigned int R() {
  static unsigned int seed = 613;
  return seed ^= seed >> 5, seed ^= seed << 17, seed ^= seed >> 13;
}
const int N = 66666;
const long long inf = 1e18;
int n;
long long l, ans1[N], ans2[N], piece;
struct node {
  int idx;
  long long l, r;
};
map<long long, long long> Map[N];
long long query(int i, long long x) {
  if (Map[i].count(x)) return Map[i][x];
  printf("? %d %lld\n", i, x);
  fflush(stdout);
  long long y;
  read(y);
  return Map[i][x] = y;
}
long long find(int i, long long val, long long l, long long r) {
  while (l < r) {
    long long mid = (l + r) >> 1, x = query(i, mid);
    if (x < val)
      l = mid + 1;
    else
      r = mid;
  }
  return l;
}
void solve(vector<node> a, long long l, long long r) {
  assert(l <= r);
  if (((int)a.size()) == 1) {
    ans1[a.back().idx] = l;
    ans2[a.back().idx] = r;
    return;
  }
  vector<node> zuo, you;
  int mid = ((int)a.size()) / 2, tot = ((int)a.size());
  long long MID = l;
  while (((int)a.size())) {
    int u = R() % ((int)a.size());
    long long p =
        find(a[u].idx, (a[u].l + (a[u].r - a[u].l) / tot * mid), l, r);
    MID = p;
    vector<node> x, y, z;
    int rec = ((int)zuo.size());
    for (register int i = (0); i <= (((int)a.size()) - 1); i++)
      if (i != u) {
        if ((a[i].l + (a[i].r - a[i].l) / tot * mid) <= query(a[i].idx, p)) {
          if (query(a[i].idx, p - 1) < (a[i].l + (a[i].r - a[i].l) / tot * mid))
            y.push_back(a[i]);
          else
            x.push_back(a[i]);
        } else
          z.push_back(a[i]);
      } else
        y.push_back(a[i]);
    int s1 = ((int)x.size()), s2 = ((int)y.size()), s3 = ((int)z.size());
    if (rec + s1 <= mid) zuo.insert(zuo.end(), x.begin(), x.end()), x.clear();
    if (rec >= mid) you.insert(you.end(), x.begin(), x.end()), x.clear();
    for (register int i = (0); i <= (((int)y.size()) - 1); i++)
      if (rec + s1 + i + 1 <= mid)
        zuo.push_back(y[i]);
      else
        you.push_back(y[i]);
    if (rec + s1 + s2 + s3 <= mid)
      zuo.insert(zuo.end(), z.begin(), z.end()), z.clear();
    if (rec + s1 + s2 >= mid)
      you.insert(you.end(), z.begin(), z.end()), z.clear();
    a = x;
    a.insert(a.end(), z.begin(), z.end());
  }
  assert(((int)zuo.size()) == mid);
  for (register int i = (0); i <= (((int)zuo.size()) - 1); i++)
    zuo[i].r = query(zuo[i].idx, MID);
  for (register int i = (0); i <= (((int)you.size()) - 1); i++)
    you[i].l = query(you[i].idx, MID);
  solve(zuo, l, MID);
  solve(you, MID, r);
}
int main() {
  read(n);
  read(l);
  piece = l / n;
  vector<node> a;
  for (register int i = (1); i <= (n); i++) a.push_back((node){i, 0, l});
  solve(a, 0, inf);
  puts("!");
  for (register int i = (1); i <= (n); i++)
    printf("%lld %lld\n", ans1[i], ans2[i]);
  fflush(stdout);
  return 0;
}