Skier rides on a snowy field. Its movements can be described by a string of characters 'S', 'N', 'W', 'E' (which correspond to 1 meter movement in the south, north, west or east direction respectively).

It is known that if he moves along a previously unvisited segment of a path (i.e. this segment of the path is visited the first time), then the time of such movement is 5 seconds. If he rolls along previously visited segment of a path (i.e., this segment of the path has been covered by his path before), then it takes 1 second.

Find the skier's time to roll all the path.

Input

The first line contains an integer t (1 ≤ t ≤ 10^4) — the number of test cases in the input. Then t test cases follow.

Each set is given by one nonempty string of the characters 'S', 'N', 'W', 'E'. The length of the string does not exceed 10^5 characters.

The sum of the lengths of t given lines over all test cases in the input does not exceed 10^5.

Output

For each test case, print the desired path time in seconds.

Example

Input


5
NNN
NS
WWEN
WWEE
NWNWS


Output


15
6
16
12
25
#include <bits/stdc++.h>
using namespace std;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t;
  cin >> t;
  while (t--) {
    string s;
    cin >> s;
    set<pair<long long int, long long int> > cow, coe, con, cos;
    pair<long long int, long long int> ccurr = make_pair(0, 0);
    long long int time = 0;
    pair<long long int, long long int> d;
    for (long long int i = 0; i < s.size(); i++) {
      pair<long long int, long long int> temp = ccurr;
      if (s[i] == 'N') {
        ccurr.second++;
      } else if (s[i] == 'S') {
        ccurr.second--;
      } else if (s[i] == 'E') {
        ccurr.first++;
      } else if (s[i] == 'W') {
        ccurr.first--;
      }
      if (s[i] == 'N' && (cos.count(ccurr) || con.count(temp))) {
        time += 1;
      } else if (s[i] == 'S' && (con.count(ccurr) || cos.count(temp))) {
        time += 1;
      } else if (s[i] == 'W' && (coe.count(ccurr) || cow.count(temp))) {
        time += 1;
      } else if (s[i] == 'E' && (coe.count(temp) || cow.count(ccurr))) {
        time += 1;
      } else {
        time += 5;
        if (s[i] == 'N') {
          con.insert(temp);
        } else if (s[i] == 'S') {
          cos.insert(temp);
        } else if (s[i] == 'E') {
          coe.insert(temp);
        } else if (s[i] == 'W') {
          cow.insert(temp);
        }
      }
    }
    cout << time << endl;
  }
}
