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 double pi = acos(-1.0);
const double eps = 1e-11;
template <class T>
inline void ckmin(T &a, T b) {
  if (b < a) a = b;
}
template <class T>
inline void ckmax(T &a, T b) {
  if (b > a) a = b;
}
template <class T>
inline T sqr(T x) {
  return x * x;
}
using VI = vector<int>;
const int MAXN = 2048;
const long long SIZE = 1000000000LL * 1000000000LL;
int n;
long long L;
pair<long long, long long> ret[MAXN];
map<long long, long long> cache[MAXN];
long long real_query(int idx, long long pos) {
  printf("? %d %lld\n", idx + 1, pos);
  fflush(stdout);
  long long val;
  scanf("%lld", &val);
  return val;
}
long long query(int idx, long long pos) {
  auto it = cache[idx].find(pos);
  if (it != cache[idx].end()) return it->second;
  return cache[idx][pos] = real_query(idx, pos);
}
bool is_greator_or_equal(int idx, long long pos, long long limit) {
  auto &w = cache[idx];
  if (!w.count(pos)) {
    auto it = w.lower_bound(pos);
    if (it != w.end()) {
      if (it->second < limit) return false;
      if (it->second - abs(pos - it->first) >= limit) return true;
    }
    if (it != w.begin()) {
      --it;
      if (it->second >= limit) return true;
      if (it->second + abs(pos - it->first) < limit) return false;
    }
  }
  return query(idx, pos) >= limit;
}
void solve(long long sx, long long tx, long long sL, long long tL, VI a) {
  assert(tL - sL == ((int)a.size()));
  if (tL - sL == 1) {
    ret[a[0]] = make_pair(sx, tx);
    return;
  }
  long long mL = (sL + tL) / 2;
  long long low = sx - 1, high = tx;
  for (; low + 1 < high;) {
    long long mid = (low + high) / 2;
    int cnt = 0;
    int left_cnt = ((int)a.size());
    for (int idx : a) {
      if (is_greator_or_equal(idx, mid, L * mL)) ++cnt;
      --left_cnt;
      if (cnt >= ((int)a.size()) / 2) break;
      if (cnt + left_cnt < ((int)a.size()) / 2) break;
    }
    if (cnt >= ((int)a.size()) / 2)
      high = mid;
    else
      low = mid;
  }
  long long mx = high;
  sort(a.begin(), a.end(),
       [&](int idx1, int idx2) { return query(idx1, mx) > query(idx2, mx); });
  VI b, c;
  for (int i = 0; i < (((int)a.size())); ++i)
    if (i < mL - sL)
      b.push_back(a[i]);
    else
      c.push_back(a[i]);
  solve(sx, mx, sL, mL, b);
  solve(mx, tx, mL, tL, c);
}
int main() {
  scanf("%d%lld", &n, &L);
  L /= n;
  VI a;
  for (int i = 0; i < (n); ++i) a.push_back(i);
  solve(0, SIZE, 0, n, a);
  random_shuffle(a.begin(), a.end());
  printf("!\n");
  for (int i = 0; i < (n); ++i)
    printf("%lld %lld\n", ret[i].first, ret[i].second);
  fflush(stdout);
  return 0;
}