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 int power(long long int a, long long int n) {
  if (n == 0) return 1;
  if ((n % 2) == 0)
    return (power((a * a), n / 2));
  else
    return (a * (power((a * a), n / 2)));
}
int main() {
  std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long int t;
  cin >> t;
  while (t--) {
    string s;
    cin >> s;
    set<pair<pair<long long int, long long int>,
             pair<long long int, long long int>>>
        b;
    long long int x = 0, y = 0, px = 0, py = 0, a = 0;
    for (int i = 0; i < s.length(); i++) {
      if (s[i] == 'N')
        y++;
      else if (s[i] == 'S')
        y--;
      else if (s[i] == 'E')
        x++;
      else
        x--;
      if ((b.find({{px, py}, {x, y}}) != b.end()) ||
          (b.find({{x, y}, {px, py}}) != b.end()))
        a++;
      else {
        a = a + 5;
        b.insert({{x, y}, {px, py}});
        b.insert({{px, py}, {x, y}});
      }
      px = x;
      py = y;
    }
    cout << a << "\n";
    b.clear();
  }
  return 0;
}
