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 MOD = 1e9 + 7;
struct star {
  int fx, fy, tx, ty;
};
bool operator<(const star& t1, const star& t2) { return (t1.fx < t2.fx); }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL), cout.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    map<pair<int, int>, map<pair<int, int>, int>> ma;
    string s;
    cin >> s;
    long long int res = 0;
    int x = 0, y = 0;
    int px = 0, py = 0;
    star p;
    for (char i : s) {
      if (i == 'N')
        y++;
      else if (i == 'S')
        y--;
      else if (i == 'W')
        x--;
      else if (i == 'E')
        x++;
      if (ma.find({px, py}) != ma.end() &&
          ma[{px, py}].find(make_pair(x, y)) != ma[{px, py}].end()) {
        res += 1;
      } else if (ma.find({x, y}) != ma.end() &&
                 ma[{x, y}].find(make_pair(px, py)) != ma[{x, y}].end()) {
        res += 1;
      } else
        res += 5, ma[{px, py}][make_pair(x, y)]++;
      px = x, py = y;
    }
    cout << res << endl;
  }
  return 0;
}
