A1008.Tests for Haybales--Gold

Farmer John's cows have decided to offer a programming contest for the cows on
Farmer Nhoj's farm. In order to make the problems as fun as possible, they
have spent considerable time coming up with challenging input cases. For one
problem in particular, "Haybales", the cows need your help devising
challenging inputs. This involve solving the following somewhat intriguing
problem:
There is an array of sorted integers $x_1 \leq x_2 \leq \dotsb \leq x_N$ ($1 \leq N \leq 10^5$), and an integer $K$. You don't know the array or $K$, but
you do know for each index $i$, the largest index $j_i$ such that $x_{j_i} \leq x_i + K$. It is guaranteed that $i\le j_i$ and $j_1\le j_2\le \cdots \le j_N\le N$.
Given this information, Farmer John's cows need to construct any array along
with some integer $K$ that matches that information. The construction needs to
satisfy $0 \leq x_i \leq 10^{18}$ for all $i$ and $1 \leq K \leq 10^{18}$.
It can be proven that this is always possible. Help Farmer John's cows solve
this problem!

The first line of input contains $N$. The next line contains
$j_1,j_2,\ldots,j_N$.

Print $K$, then $x_1,\ldots,x_N$ on separate lines. Any valid output will be
accepted.

The sample output is the array $a = [1, 6, 17, 22, 27, 32]$ with $K = 6$. $j_1 = 2$ is satisfied because $a_2 = 6 \leq 1 + 6 = a_1 + K$ but $a_3 = 17 > 1 + 6 = a_1 + K$, so $a_2$ is the largest element that is at most $a_1$. Similarly,

• $j_2 = 2$ is satisfied because $a_2 = 6 \leq 6 + 6$ but $a_3 = 17 > 6 + 6$
• $j_3 = 4$ is satisfied because $a_4 = 22 \leq 17 + 6$ but $a_5 = 27 > 17 + 6$
• $j_4 = 5$ is satisfied because $a_5 = 27 \leq 22 + 6$ but $a_5 = 32 > 22 + 6$
• $j_5 = 6$ is satisfied because $a_6 = 32 \leq 27 + 6$ and $a_6$ is the last element of the array
• $j_6 = 6$ is satisfied because $a_6 = 32 \leq 32 + 6$ and $a_6$ is the last element of the array
  This is not the only possible correct output for the sample input. For
  example, you could instead output the array $[1, 2, 4, 5, 6, 7]$ with $K = 1$.