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 int N = 1005;
long long a[N][N];
bool type;
int n, Qcnt;
long long V;
long long rnd() {
  long long x = 0;
  for (int i = (int)(1); i <= (int)(18); i++) x = x * 10 + rand() % 10;
  return x;
}
void init() {
  type = 1;
  for (int i = (int)(1); i <= (int)(n); i++) {
    int top = 0;
    for (; top != V;) {
      for (; top != V; a[i][++top] = rnd())
        ;
      sort(a[i] + 1, a[i] + V + 1);
      top = unique(a[i] + 1, a[i] + V + 1) - a[i] - 1;
    }
  }
}
long long query(int x, long long y) {
  ++Qcnt;
  if (type == 0) {
    printf("? %d %lld\n", x, y);
    fflush(stdout);
    scanf("%lld", &y);
    return y;
  } else {
    int pos = upper_bound(a[x] + 1, a[x] + V * n + 1, y) - a[x] - 1;
    return pos;
  }
}
int id[N], idd[N];
long long val[N], ansR[N], ansL[N];
bool cmp(int x, int y) { return val[x] < val[y]; }
map<pair<long long, long long>, long long> mp;
long long find(int x, long long L, long long R, long long val) {
  if (mp.find(pair<long long, long long>(x, val)) != mp.end())
    return mp[pair<long long, long long>(x, val)];
  long long ans = 0;
  for (; L <= R;) {
    long long mid = (L + R) / 2;
    long long v = query(x, mid);
    if (v == val) ans = mid;
    if (v >= val)
      R = mid - 1;
    else
      L = mid + 1;
  }
  return mp[pair<long long, long long>(x, val)] = ans;
}
void divide(long long L, long long R, long long l, long long r) {
  if (l > r) return;
  if (L == R) {
    for (int i = (int)(l); i <= (int)(r); i++) ansR[id[i]] = L;
    return;
  }
  if (l == r) {
    ansR[id[l]] = find(id[l], L, R, l * V);
    assert(ansR[id[l]] != 0);
    return;
  }
  int mid = (l + r) / 2;
  int pl = l, pr = r;
  long long LL = L, RR = R;
  for (;;) {
    int speP = id[pl + rand() % (pr - pl + 1)];
    long long pos = find(speP, LL, RR, mid * V);
    int plll = pl, prrr = pr;
    for (int i = (int)(pl); i <= (int)(pr); i++) {
      long long val = query(id[i], pos);
      if (val > mid * V || (val == mid * V && id[i] <= speP))
        idd[plll++] = id[i];
      else
        idd[prrr--] = id[i];
    }
    for (int i = (int)(pl); i <= (int)(pr); i++) id[i] = idd[i];
    if (prrr <= mid)
      pl = prrr + 1, LL = pos;
    else
      pr = plll - 1, RR = pos;
    if (pr <= mid || pl > mid) break;
  }
  divide(L, RR, l, mid);
  divide(LL, R, mid + 1, r);
}
void output() {
  for (int i = (int)(1); i <= (int)(n); i++) ansL[id[i]] = ansR[id[i - 1]];
  printf("!\n");
  for (int i = (int)(1); i <= (int)(n); i++)
    printf("%lld %lld\n", ansL[i], ansR[i]);
  fflush(stdout);
}
int main() {
  scanf("%d%lld", &n, &V);
  srand(time(NULL));
  V /= n;
  for (int i = (int)(1); i <= (int)(n); i++) id[i] = i;
  divide(0, 1000000000000000000ll, 1, n);
  output();
}