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

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.

Example

0010100

3
3 1 3 4
3 2 5 6
1 7

111

-1

超級大模擬
碰到0，優先加入1結尾的數列，不然新建一個數列
碰到1，加入以0結尾的數列，如果沒有以0結尾的數列，那麽就不存在可xing的分配情況
最後再檢測是否全部數列都以0結尾，如果存在以1結尾的數列那麽不存在可行的分配情況
另外需要註意的是，建立一個1結尾隊列索引和0結尾隊列索引的數列，不需要每次去真的
找一遍0結尾和1結尾的隊列

代碼:

 1 #include<iostream>
 2 using namespace std;
 3 #include<cstdio>
 4 #include<cstring>
 5 #include<vector>
 6 #include<cstdlib>
 7 const int maxn = 1e5*5;
 8 char s[maxn];
 9 vector<int> a[maxn];
10 vector<int> v0;
11 vector<int> v1;
12 int num = 0;
13 int main(){
14     scanf("%s",s);
15     int ok=1;
16     int len =strlen(s);
17     int pt;
18     for(int i=0;i<len;i++){
19         s[i] = s[i] - ‘0‘;
20         if(s[i]==0){
21             if(v1.size()==0){
22                 v0.push_back(num);
23                 a[num].push_back(i+1);
24                 num+=1;
25             }
26             else{
27                 pt = v1[v1.size()-1];
28                 v1.pop_back();
29                 v0.push_back(pt);
30                 a[pt].push_back(i+1);
31             }
32         }
33         else if(s[i]==1){
34             if(v0.size()==0){
35                 ok=0;
36                 break;
37             }
38             else{
39                 int pt = v0[v0.size()-1];
40                 v0.pop_back();
41                 v1.push_back(pt);
42                 a[pt].push_back(i+1);
43             }
44         }
45     }
46     if(v1.size()!=0)
47         ok = 0;
48     if(ok==0){
49         printf("-1");
50     }
51     else{
52         printf("%d\n",num);
53         for(int i=0;i<num;i++){
54             printf("%d",a[i].size());
55             for(int j=0;j<a[i].size();j++){
56                 printf(" %d",a[i][j]);
57             }
58             putchar(‘\n‘);
59         }
60     }
61     return 0;
62 }

CodeForces - 950C Zebras VirtualJudge 500題AC留念