Problem: You are given a non-empty string s=s_1s_2... s_n, 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 s_{j}s_{j+1}s_{j+2}="one" or s_{j}s_{j+1}s_{j+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 ≤ 10^4) — 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⋅10^5. 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⋅10^6.

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 

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". 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const long long int MOD = 1000000000 + 7;
const long long int MOD1 = 998244353;
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long int strTo(string s) {
  stringstream tmp(s);
  long long int x = 0;
  tmp >> x;
  return x;
}
string NoTo(long long int n) {
  ostringstream str1;
  str1 << n;
  string ans = str1.str();
  return ans;
}
vector<bool> sieve(1000000, true);
void Sieve() {
  sieve[0] = false;
  sieve[1] = false;
  for (long long int i = 2; i * i <= 1000000; i++) {
    if (sieve[i] == true) {
      for (long long int j = i * i; j < 1000000; j += i) sieve[j] = false;
    }
  }
}
long long int lcm(long long int a, long long int b) {
  return (a * b) / gcd(a, b);
}
long long int arrlcm(long long int arr[], long long int n) {
  long long int ans = arr[0];
  for (long long int i = 1; i < n; i++)
    ans = (((arr[i] * ans)) / (gcd(arr[i], ans)));
  return ans;
}
long long int arrgcd(long long int arr[], long long int n) {
  long long int result = arr[0];
  for (long long int i = 1; i < n; i++) {
    result = gcd(arr[i], result);
    if (result == 1) {
      return 1;
    }
  }
  return result;
}
long long int nCr(long long int n, long long int r) {
  long long int ans = 1;
  for (long long int i = (long long int)0; i < (long long int)r; i++) {
    ans *= n - i;
    ans /= i + 1;
  }
  return ans;
}
long long int fact(long long int n) {
  if (n == 0) return 1;
  return n * fact(n - 1);
}
vector<long long int> v;
void Divisors(long long int n) {
  for (long long int i = 1; i <= sqrt(n); i++) {
    if (n % i == 0) {
      if (n / i == i)
        v.push_back(i);
      else {
        v.push_back(i);
        v.push_back(n / i);
      }
    }
  }
}
long long int no_of_divisors(long long int n) {
  long long int c = 0;
  for (long long int i = 1; i <= sqrt(n); i++) {
    if (n % i == 0) {
      if (n / i == i)
        c = c + 1;
      else
        c = c + 2;
    }
  }
  return c;
}
long long int pwrn(long long int x, long long int y) {
  if (y == 0)
    return 1;
  else if (y % 2 == 0)
    return pwrn(x, y / 2) * pwrn(x, y / 2);
  else
    return x * pwrn(x, y / 2) * pwrn(x, y / 2);
}
long long int pwr(long long int a, long long int b) {
  long long int res = 1;
  while (b > 0) {
    if (b & 1) res = res * a;
    a *= a;
    b >>= 1;
  }
  return res;
}
long long int pwrm(long long int a, long long int b, long long int m) {
  a %= m;
  long long int res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % m;
    a = a * a % m;
    b >>= 1;
  }
  return res;
}
long long int digits(long long int n) {
  long long int x = floor(log10(n)) + 1;
  return x;
}
vector<long long int> prf;
void primeFactors(long long int n) {
  if (n % 2 == 0) prf.push_back(2);
  while (n % 2 == 0) {
    n = n / 2;
  }
  for (long long int i = 3; i <= sqrt(n); i = i + 2) {
    if (n % i == 0) prf.push_back(i);
    while (n % i == 0) {
      n = n / i;
    }
  }
  if (n > 1) prf.push_back(n);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long int T;
  cin >> T;
  while (T--) {
    string s;
    cin >> s;
    string t;
    t += s[0];
    t += s[1];
    long long int ans = 0;
    vector<long long int> v;
    for (long long int i = (long long int)2; i < (long long int)s.size(); i++) {
      if (s[i] == 'e') {
        if (t[t.size() - 1] == 'n' && t[t.size() - 2] == 'o') {
          v.push_back(i);
          t.pop_back();
          t += 'e';
        } else {
          t += 'e';
        }
      } else if (s[i] == 'o') {
        if (t[t.size() - 1] == 'w' && t[t.size() - 2] == 't' &&
            (i + 1) < s.size() && (i + 2) < s.size() && s[i + 1] == 'n' &&
            s[i + 2] == 'e') {
          v.push_back(i + 1);
        } else if (t[t.size() - 1] == 'w' && t[t.size() - 2] == 't') {
          v.push_back(i);
          t.pop_back();
          t += 'o';
        } else {
          t += 'o';
        }
      } else {
        t += s[i];
      }
    }
    cout << v.size() << "\n";
    for (long long int i = (long long int)0; i < (long long int)v.size(); i++)
      cout << v[i] << " ";
    cout << "\n";
  }
  cerr << "Time : " << 1000 * ((double)clock()) / (double)CLOCKS_PER_SEC
       << "ms\n";
  return 0;
}