As Simple as One and Two

You are given a non-empty string s=s1s2…sns=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 ss if there is an integer jj (1≤j≤n−21≤j≤n−2), that sjsj+1sj+2=sjsj+1sj+2="one" or sjsj+1sj+2=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=s="onetwone", then if Polycarp selects two indices 33 and 66, 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 tt (1≤t≤1041≤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 ss. Its length does not exceed 1.5⋅1051.5⋅105. The string ss 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⋅1061.5⋅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 rr (0≤r≤|s|0≤r≤|s|) — the required minimum number of positions to be removed, where |s||s| is the length of the given line. The second line of each answer should contain rr different integers — the indices themselves for removal in any order. Indices are numbered from left to right from 11 to the length of the string. If r=0r=0, then the second line can be skipped (or you can print empty). If there are several answers, print any of them.

Examples

input

Copy

4
onetwone
testme
oneoneone
twotwo

output

Copy

2
6 3
0

3
4 1 7 
2
1 4

input

Copy

10
onetwonetwooneooonetwooo
two
one
twooooo
ttttwo
ttwwoo
ooone
onnne
oneeeee
oneeeeeeetwooooo

output

Copy

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

1
4 
0

1
1 
2
1 11 

Note

In the first example, answers are:

• "onetwone",
• "testme" — Polycarp likes it, there is nothing to remove,
• "oneoneone",
• "twotwo".

In the second example, answers are:

• "onetwonetwooneooonetwooo",
• "two",
• "one",
• "twooooo",
• "ttttwo",
• "ttwwoo" — Polycarp likes it, there is nothing to remove,
• "ooone",
• "onnne" — Polycarp likes it, there is nothing to remove,
• "oneeeee",
• "oneeeeeeetwooooo".

---

感觉挺迷惑的一道题。。。

#include <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <set>
#include <queue>
#include <map>
#include <sstream>
#include <cstdio>
#include <cstring>
#include <numeric>
#include <cmath>
#include <unordered_set>
#include <unordered_map>
#include <xfunctional>
#define ll long long
#define mod 998244353
using namespace std;
int dir[4][2] = { {0,1},{0,-1},{-1,0},{1,0} };
const int maxn = 1e5 + 5;
const long long inf = 0x7f7f7f7f7f7f7f7f;

int main()
{
    int t;
    cin >> t;
    while (t--)
    {
        string s;
        cin >> s;
        vector<int> v;
        for (int i = 0; i < s.size(); i++)
        {
            if (s[i] == 't')
            {
                i++;
                if (i < s.size() && s[i] == 'w')
                {
                    i++;
                    if (i < s.size() && s[i] == 'o')
                    {
                        i++;
                        if (i < s.size() && s[i] != 'o')
                            v.push_back(i - 1);
                        else
                        {
                            v.push_back(i - 2);
                        }
                        i--;
                        continue;
                    }
                    i--;
                    continue;
                }
                i--;
                continue;
            }
            if (s[i] == 'o')
            {
                i++;
                if (i < s.size() && s[i] == 'n')
                {
                    i++;
                    if (i < s.size() && s[i] == 'e')
                    {                        
                        if (i-3>=0 && s[i - 3] == 'o')
                            v.push_back(i - 1);
                        else
                            v.push_back(i - 2);
                    }
                    i--;
                }
                i--;
            }
        }
        cout << v.size() << endl;
        if (v.size() == 0)
            cout << endl;
        for (int i = 0; i < v.size(); i++)
        {
            cout << v[i] + 1 << " ";
            if (i == v.size() - 1)
                cout << endl;
        }
    }

    return 0;
}