As Simple as One and Two

time limit per test3 seconds
memory limit per test256 megabytes
input: standard input
output: standard output

You are given a non-empty string s=s1s2…sn, which consists only of lowercase Latin letters. Polycarp does not like a string if it contains at least one string “one” or at least one string “two” (or both at the same time) as a substring. In other words, Polycarp does not like the string s if there is an integer j(1≤j≤n−2), that sjsj+1sj+2=“one” or sjsj+1sj+2=“two”.
For example:

Polycarp does not like strings “oneee”, “ontwow”, “twone” and “oneonetwo” (they all have at least one substring “one” or “two”),
Polycarp likes strings “oonnee”, “twwwo” and “twnoe” (they have no substrings “one” and “two”).

Polycarp wants to select a certain set of indices (positions) and remove all letters on these positions. All removals are made at the same time.
For example, if the string looks like s=“onetwone”, then if Polycarp selects two indices 3 and 6, then “onetwone” will be selected and the result is “ontwne”.
What is the minimum number of indices (positions) that Polycarp needs to select to make the string liked? What should these positions be?

Input
The first line of the input contains an integer t(1≤t≤104) — the number of test cases in the input. Next, the test cases are given.
Each test case consists of one non-empty string s. Its length does not exceed 1.5⋅105. The string s consists only of lowercase Latin letters.
It is guaranteed that the sum of lengths of all lines for all input data in the test does not exceed 1.5⋅106.

Output
Print an answer for each test case in the input in order of their appearance.
The first line of each answer should contain r(0≤r≤|s|) — the required minimum number of positions to be removed, where |s| is the length of the given line. The second line of each answer should contain r different integers — the indices themselves for removal in any order. Indices are numbered from left to right from 1 to the length of the string. If r=0, then the second line can be skipped (or you can print empty). If there are several answers, print any of them.

Examples

Input
4
onetwone
testme
oneoneone
twotwo

Output
2
6 3
0
3
4 1 7
2
1 4

Input

10
onetwonetwooneooonetwooo
two
one
twooooo
ttttwo
ttwwoo
ooone
onnne
oneeeee
oneeeeeeetwooooo

Output
6
18 11 12 1 6 21
1
1
1
3
1
2
1
6
0
1
4
0
1
1
2
1 11

Title effect:
Given a string, delete part of (or may not be deleted), so that no continuous character "one" or "two".
Cf c title of a recent field div2, strategies, there is some question initially, leading to even wa three rounds, instant explosion penalty. Strategy is simple, if a separate "one" n put deleted, if a separate "two" put w deleted, can make similar "oooneee" with such a case where the head end of the plurality of characters. And if it is "twone", put o deleted. The first wa determination is not done because, for loop pointer does not jump the "twone" o deleted after the loop to the "one" n again to delete. After the addition of a logo arrays, honey wa, finally made directly to the cursor jumps under treatment on the ac. (Do not just hit the brain is not clear title .jpg)

Code

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
int t;
string s;
int a[100005];
int times,num;
 
int main () {
	cin>>t;
	while(t--){
		cin>>s;
		times=0;	num=0;
		for(int i=0;i<s.size();i++){
			if(s[i]=='o' && s[i+1]=='n'&& s[i+2]=='e'){
				times++;
				a[num++]=i+2;
				i=i+2;
			}else if(s[i]=='t' && s[i+1]=='w' && s[i+2]=='o'){
				if(s[i+3]=='n' && s[i+4]=='e'){
					times++;
					a[num++]=i+3;
					i=i+4;
				}else{
					times++;
					a[num++]=i+2;
					i=i+2;
				}	
			}
		}
		cout<<times<<endl;
		for(int i=0;i<num;i++){
			cout<<a[i];
			if(i!=num-1){
				cout<<" ";
			}
		}
		cout<<endl;
	}
	return 0;
	
}

　　

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CSDN link: https: //blog.csdn.net/weixin_43880627/article/details/103621781