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() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t, x, y, ans, prevx, prevy;
  string s;
  cin >> t;
  while (t--) {
    cin >> s;
    x = 0;
    y = 0;
    prevx = 0;
    prevy = 0;
    ans = 0;
    set<pair<pair<int, int>, pair<int, int> > > st;
    for (int i = 0; i < s.size(); i++) {
      if (s[i] == 'N')
        y += 1;
      else if (s[i] == 'S')
        y -= 1;
      if (s[i] == 'E')
        x += 1;
      else if (s[i] == 'W')
        x -= 1;
      pair<int, int> e1(x, y);
      pair<int, int> e2(prevx, prevy);
      pair<pair<int, int>, pair<int, int> > e3(e1, e2);
      if (st.find(e3) == st.end()) {
        ans += 5;
        st.insert(e3);
        st.insert(pair<pair<int, int>, pair<int, int> >(e2, e1));
        prevx = x;
        prevy = y;
      } else {
        ans += 1;
        prevx = x;
        prevy = y;
      }
    }
    cout << ans << endl;
  }
  return 0;
}
