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--) {
    int time = 0, x = 0, y = 0, i, n, x1 = 0, y1 = 0;
    double xmid = 0, ymid = 0;
    string s;
    cin >> s;
    n = s.length();
    set<pair<int, int> > st;
    set<pair<double, double> > st1;
    st.insert(make_pair(x, y));
    for (i = 0; i < n; i++) {
      if (s[i] == 'N') {
        y++;
      }
      if (s[i] == 'S') {
        y--;
      }
      if (s[i] == 'W') {
        x++;
      }
      if (s[i] == 'E') {
        x--;
      }
      xmid = (double)(x1 + x) / 2.0;
      ymid = (double)(y1 + y) / 2.0;
      if ((st.find(make_pair(x, y)) == st.end()) ||
          (st1.find(make_pair(xmid, ymid)) == st1.end())) {
        time += 5;
      } else {
        time++;
      }
      st.insert(make_pair(x, y));
      st1.insert(make_pair(xmid, ymid));
      x1 = x;
      y1 = y;
    }
    cout << time << endl;
  }
  return 0;
}
