CF949A.Zebras

Oleg writes down the history of the days he lived. For each day he decides if it was good or bad. Oleg calls a non-empty sequence of days a zebra, if it starts with a bad day, ends with a bad day, and good and bad days are alternating in it. Let us denote bad days as 0 and good days as 1. Then, for example, sequences of days 0, 010, 01010 are zebras, while sequences 1, 0110, 0101 are not.

Oleg tells you the story of days he lived in chronological order in form of string consisting of 0 and 1. Now you are interested if it is possible to divide Oleg's life history into several subsequences, each of which is a zebra, and the way it can be done. Each day must belong to exactly one of the subsequences. For each of the subsequences, days forming it must be ordered chronologically. Note that subsequence does not have to be a group of consecutive days.

奥列格记录下自己所经历的每一天的历史。对于每一天，他都会判断这一天是“好”的还是“坏”的。奥列格将一个非空的天数序列称为“斑马序列”，当且仅当：该序列以“坏”天开始、以“坏”天结束，且其中“好”天与“坏”天严格交替出现。我们用 0 表示“坏”天，用 1 表示“好”天。例如，序列 0、010、01010 是斑马序列；而序列 1、0110、0101 则不是。

奥列格按时间顺序向你讲述他所经历的日子，其形式为一个仅由 0 和 1 组成的字符串。现在你想知道：是否可以将奥列格的生活历史划分为若干个子序列，使得每个子序列都是一个斑马序列？若可以，请给出一种划分方式。要求：每一天恰好属于且仅属于其中一个子序列；且对每个子序列而言，其所包含的天数必须保持其原始的时间顺序（即子序列中各天在原字符串中的相对顺序不变）。注意：子序列中的天数不必连续。

In the only line of input data there is a non-empty string s consisting of characters 0 and 1, which describes the history of Oleg's life. Its length (denoted as |s|) does not exceed 200 000 characters.

输入数据仅有一行，包含一个非空字符串 $s$，该字符串由字符 0 和 1 组成，用于描述奥列格的人生历程。其长度（记为 $|s|$）不超过 $200\,000$ 个字符。

If there is a way to divide history into zebra subsequences, in the first line of output you should print an integer k (1 ≤ k ≤ |s|), the resulting number of subsequences. In the i-th of following k lines first print the integer l__i (1 ≤ l__i ≤ |s|), which is the length of the i-th subsequence, and then l__i indices of days forming the subsequence. Indices must follow in ascending order. Days are numbered starting from 1. Each index from 1 to n must belong to exactly one subsequence. If there is no way to divide day history into zebra subsequences, print -1.

Subsequences may be printed in any order. If there are several solutions, you may print any of them. You do not have to minimize nor maximize the value of k.

如果可以将历史划分为斑马子序列，则在输出的第一行中，应打印一个整数 $k$（$1 ≤ k ≤ |s|$），即最终得到的子序列数量。在接下来的 $k$ 行中，第 $i$ 行首先打印整数 $l_i$（$1 ≤ l_i ≤ |s|$），表示第 $i$ 个子序列的长度，然后打印 $l_i$ 个构成该子序列的日期下标。下标必须按升序排列。日期编号从 1 开始。每个从 1 到 $n$ 的下标必须恰好属于一个子序列。如果无法将日期历史划分为斑马子序列，则输出 $-1$。

子序列的输出顺序可以任意。若存在多种解法，可输出其中任意一种。你无需最小化或最大化 $k$ 的值。