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;
using ll = long long;
const int N = 1005;
int a[N], b[N], c[N];
ll n, len, val;
pair<ll, ll> ans[N];
inline ll query(ll a, ll b) {
  cout << "? " << a << ' ' << b << endl;
  ll ret;
  cin >> ret;
  return ret;
}
ll find(int x, ll kk, ll l, ll r) {
  ll mid = (l + r) >> 1;
  while (1) {
    ll y = query(x, mid);
    if (y == kk) return mid;
    if (y < kk)
      l = mid + 1;
    else
      r = mid - 1;
    mid = (l + r) >> 1;
  }
}
void kth(int l, int r, int kk, ll ql, ll qr) {
  if (l == r) return;
  int x = rand() % (r - l + 1) + l;
  ll y = find(a[x], (len / n) * kk, ql, qr);
  int l1 = l - 1, l2 = r + 1, l3 = 0;
  c[++l3] = a[x];
  for (int i = l; i <= r; i++) {
    if (i == x) continue;
    ll tt = query(a[i], y);
    if (tt < (len / n) * kk) b[--l2] = a[i];
    if (tt == (len / n) * kk) c[++l3] = a[i];
    if (tt > (len / n) * kk) b[++l1] = a[i];
  }
  while (l1 < kk && l3) b[++l1] = c[l3--];
  if (l1 == kk) {
    val = y;
    while (l3) b[--l2] = c[l3--];
    for (int i = l; i <= r; i++) a[i] = b[i];
    return;
  }
  if (l1 < kk) {
    for (int i = l; i <= r; i++) a[i] = b[i];
    return kth(l1 + 1, r, kk, ql, qr);
  }
  while (l3) b[--l2] = c[l3--];
  for (int i = l; i <= r; i++) a[i] = b[i];
  return kth(l, l1, kk, ql, qr);
}
void solve(int l, int r, ll ql, ll qr) {
  if (l == r) {
    ans[a[l]] = {ql, qr};
    return;
  }
  int mid = (l + r) >> 1;
  kth(l, r, mid, ql, qr);
  ll x = val;
  solve(l, mid, ql, x);
  solve(mid + 1, r, x, qr);
}
int main() {
  ios::sync_with_stdio(false);
  srand(time(0));
  cin >> n >> len;
  for (int i = 1; i <= n; i++) a[i] = i;
  solve(1, n, 0, 1000000000000000000);
  cout << "!\n";
  for (int i = 1; i <= n; i++)
    cout << ans[i].first << ' ' << ans[i].second << endl;
  return 0;
}