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;
long long power(long long a, long long b) {
  if (!b) return 1;
  long long temp = power(a, b / 2);
  if (b & 1) return (temp * temp) * a;
  return (temp * temp);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t = 1;
  cin >> t;
  while (t--) {
    string str;
    cin >> str;
    set<tuple<long long, long long, long long, long long>> paths;
    long long x = 0, y = 0, time = 0;
    for (long long i = 0; i < str.size(); ++i) {
      long long x1 = x, y1 = y;
      switch (str[i]) {
        case 'N':
          y1++;
          break;
        case 'S':
          y1--;
          break;
        case 'W':
          x1++;
          break;
        case 'E':
          x1--;
          break;
      }
      tuple<long long, long long, long long, long long> new_path = {x1, y1, x,
                                                                    y},
                                                        new_path2 = {x, y, x1,
                                                                     y1};
      if (paths.count(new_path2) || paths.count(new_path))
        time++;
      else
        time += 5;
      paths.insert(new_path);
      paths.insert(new_path2);
      x = x1;
      y = y1;
    }
    cout << time << "\n";
  }
  return 0;
}
