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 SIZE = (long long)1e18;
int remains = 200000;
vector<set<pair<long long, long long>>> evals;
long long need;
vector<long long> resLeft;
vector<long long> resRight;
long long realEval(int fid, long long x) {
  --remains;
  assert(remains >= 0);
  printf("? %d %lld\n", (fid + 1), x);
  fflush(stdout);
  long long realY;
  scanf("%lld", &realY);
  return realY;
}
bool compareAt(int fid, long long x, long long y) {
  set<pair<long long, long long>>& known = evals[fid];
  auto next = known.lower_bound(pair<long long, long long>(x + 1, -1));
  auto prev = next;
  --prev;
  assert(prev->first >= 0);
  assert(prev->first <= x);
  assert(next->first > x);
  assert(next->first <= SIZE + 1);
  if (prev->second >= y) return true;
  if (prev->second + x - prev->first < y) return false;
  if (next->second < y) return false;
  if (next->second + x - next->first >= y) return true;
  long long realY = realEval(fid, x);
  known.insert(pair<long long, long long>(x, realY));
  return realY >= y;
}
void rec(vector<int>& ids, long long left, long long right, long long base) {
  if (ids.size() == 1) {
    resLeft[ids[0]] = left;
    resRight[ids[0]] = right;
    return;
  }
  int n = ids.size();
  int n1 = n / 2;
  long long interesting = base + n1 * need;
  long long searchLeft = 0;
  long long searchRight = SIZE;
  while (searchRight - searchLeft > 1) {
    long long searchMiddle = (searchLeft + searchRight) / 2;
    int count = 0;
    int togo = n;
    for (int x : ids) {
      if (compareAt(x, searchMiddle, interesting)) {
        ++count;
      }
      if (count >= n1) break;
      --togo;
      if (count + togo < n1) break;
    }
    if (count >= n1) {
      searchRight = searchMiddle;
    } else {
      searchLeft = searchMiddle;
    }
  }
  vector<int> ids1(n1);
  vector<int> ids2(n - n1);
  int p1 = 0;
  int p2 = 0;
  for (int x : ids) {
    if (compareAt(x, searchLeft, interesting)) {
      ids1[p1++] = x;
    } else if (!compareAt(x, searchRight, interesting)) {
      ids2[p2++] = x;
    }
  }
  for (int x : ids) {
    if (compareAt(x, searchLeft, interesting)) {
    } else if (!compareAt(x, searchRight, interesting)) {
    } else if (p1 == ids1.size()) {
      ids2[p2++] = x;
    } else {
      ids1[p1++] = x;
    }
  }
  rec(ids1, left, searchRight, base);
  rec(ids2, searchRight, right, interesting);
}
int main() {
  int n;
  scanf("%d", &n);
  scanf("%lld", &need);
  need /= n;
  for (int i = 0; i < n; ++i) {
    set<pair<long long, long long>> pts;
    pts.insert(pair<long long, long long>(0, 0));
    pts.insert(pair<long long, long long>(SIZE, need * n));
    pts.insert(pair<long long, long long>(SIZE + 1, need * n));
    evals.push_back(pts);
  }
  vector<int> ids;
  for (int i = 0; i < n; ++i) ids.push_back(i);
  resLeft.resize(n);
  resRight.resize(n);
  rec(ids, 0, SIZE, 0);
  for (int i = 0; i < n; ++i) {
    assert(realEval(i, resRight[i]) - realEval(i, resLeft[i]) >= need);
  }
  printf("!\n");
  for (int i = 0; i < n; ++i) {
    printf("%lld %lld\n", resLeft[i], resRight[i]);
  }
  fflush(stdout);
}