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;
int t;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  cin >> t;
  while (t--) {
    string s;
    cin >> s;
    pair<int, int> pos{0, 0};
    set<vector<int>> book;
    int ans = 0;
    vector<int> path;
    for (char &c : s) {
      switch (c) {
        case 'S':
          path = vector<int>{pos.first, pos.second, pos.first + 1, pos.second};
          ans += (book.count(path) ? 1 : 5);
          book.insert(
              vector<int>{pos.first, pos.second, pos.first + 1, pos.second});
          book.insert(
              vector<int>{pos.first + 1, pos.second, pos.first, pos.second});
          ++pos.first;
          break;
        case 'N':
          path = vector<int>{pos.first, pos.second, pos.first - 1, pos.second};
          ans += (book.count(path) ? 1 : 5);
          book.insert(
              vector<int>{pos.first, pos.second, pos.first - 1, pos.second});
          book.insert(
              vector<int>{pos.first - 1, pos.second, pos.first, pos.second});
          --pos.first;
          break;
        case 'W':
          path = vector<int>{pos.first, pos.second, pos.first, pos.second - 1};
          ans += (book.count(path) ? 1 : 5);
          book.insert(
              vector<int>{pos.first, pos.second, pos.first, pos.second - 1});
          book.insert(
              vector<int>{pos.first, pos.second - 1, pos.first, pos.second});
          --pos.second;
          break;
        case 'E':
          path = vector<int>{pos.first, pos.second, pos.first, pos.second + 1};
          ans += (book.count(path) ? 1 : 5);
          book.insert(
              vector<int>{pos.first, pos.second, pos.first, pos.second + 1});
          book.insert(
              vector<int>{pos.first, pos.second + 1, pos.first, pos.second});
          ++pos.second;
          break;
        default:
          break;
      }
    }
    cout << ans << endl;
  }
  cerr << "Time : " << (double)clock() / (double)CLOCKS_PER_SEC << "s\n";
  return 0;
}
