Description:
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 $$$f_{1}, \ldots, f_{n}$$$, 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 $$$r_{i}$$$ ($$$0 \leq l_{i} \leq r_{i} \leq 10^{18}$$$), such that $$$f_{i}(r_{i}) - f_{i}(l_{i}) \geq \frac{L}{n}$$$ (here $$$f_i(x)$$$ means the value of $$$i$$$-th function at point $$$x$$$) for all $$$i$$$ such that $$$1 \leq i \leq n$$$ 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 $$$2 \cdot 10^{5}$$$ 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 Format:
The first line contains two integers $$$n$$$ and $$$L$$$ ($$$1 \leq n \leq 1000$$$, $$$1 \leq L \leq 10^{18}$$$, $$$n$$$ is a divisor of $$$L$$$) — number of functions and their value in $$$10^{18}$$$.
Output Format:
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.
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 $$$\frac{L}{n}$$$ (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.