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 main() {
  int t;
  cin >> t;
  while (t--) {
    set<pair<pair<int, int>, int>> vis;
    int a = 0, b = 0, count, ans = 0, c = 0, d = 0, e = 0, f = 0;
    string s;
    cin >> s;
    for (int i = 0; i < s.length(); i++) {
      e = c;
      f = d;
      if (s[i] == 'N') {
        a = 1;
        d++;
      } else if (s[i] == 'S') {
        a = -1;
        d--;
      } else if (s[i] == 'E') {
        a = 2;
        c++;
      } else if (s[i] == 'W') {
        a = -2;
        c--;
      }
      if (vis.count(make_pair(make_pair(e, f), -1 * a)) == 0)
        vis.insert(make_pair(make_pair(c, d), a));
    }
    cout << vis.size() * 5 + s.length() - vis.size() << "\n";
  }
  return 0;
}
