F - Zebras CodeForces - 950C (vector建图

题目来源：

https://vjudge.net/contest/292642#problem/F

https://codeforces.com/problemset/problem/950/C

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.

Input

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.

Output

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 li (1 ≤ li ≤ |s|), which is the length of the i-th subsequence, and then li 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.

Examples

Input

Output

Input

Output

-1

题意：

给定一个01字符串，需要你把它分为k个子序列，其中k可以为任意正整数。

对子序列的要求为

以0开始，以0结束

输出满足条件的一种结果即可。

思路：

遇到一个0 先看有没有以1结尾的串有就补上没有就新开一个串，遇到一个1 有以0结尾的串就填上没有就输出-1。

用vector来保存每个子序列的下标集合，每次遇到0的时候可以先考虑放到最后一个元素为1的容器集合中，如果没有的话放到一个新的vector中。遇到1的时候就必须放在最后一个元素为0的容器中。然后根据这样模拟即可，详细看代码会更直观点。

参考代码：

#include <cstdio>
#include<cstring>
#include<vector>
using namespace std;
#define N 200005
char s[N];
vector<int>ans[N];
int main()
{
    scanf("%s",s+1);
    int len=strlen(s+1);
    int Max_size=0,id=0;
    for(int i=1; i<=len; i++)
    {
        if(s[i]=='0')
            ans[++id].push_back(i);//以0结尾的
        else
        {
            if(id==0)
            {
                puts("-1");//必须以0开始
                return 0;
            }
            ans[id--].push_back(i);//这个位置放了1之后下一次又可以放0了
        }
        Max_size=max(Max_size,id); //Max为当前已经用了几个容器，即分成了几个子序列
    }
    if(Max_size!=id)
    {
        puts("-1");//必须以0结尾
        return 0;
    }
    printf("%d\n",Max_size);
    for(int i=1; i<=Max_size; i++)
    {
        printf("%d",ans[i].size());
        for(int j=0; j<ans[i].size(); j++)
            printf(" %d",ans[i][j]);
        puts("");
    }
    return 0;
}