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;
  set<pair<pair<pair<int, int>, int>, int> > s;
  string str;
  while (t--) {
    cin >> str;
    int time = 0, x = 0, y = 0;
    for (int i = 0; i < str.size(); i++) {
      if (str[i] == 'N') {
        y++;
        if (s.count({{{x, y}, x}, y - 1})) {
          time++;
        } else {
          time += 5;
          s.insert({{{x, y}, x}, y - 1});
          s.insert({{{x, y - 1}, x}, y});
        }
      }
      if (str[i] == 'S') {
        y--;
        if (s.count({{{x, y}, x}, y + 1})) {
          time++;
        } else {
          time += 5;
          s.insert({{{x, y}, x}, y + 1});
          s.insert({{{x, y + 1}, x}, y});
        }
      }
      if (str[i] == 'E') {
        x++;
        if (s.count({{{x, y}, x - 1}, y})) {
          time++;
        } else {
          time += 5;
          s.insert({{{x, y}, x - 1}, y});
          s.insert({{{x - 1, y}, x}, y});
        }
      }
      if (str[i] == 'W') {
        x--;
        if (s.count({{{x, y}, x + 1}, y})) {
          time++;
        } else {
          time += 5;
          s.insert({{{x, y}, x + 1}, y});
          s.insert({{{x + 1, y}, x}, y});
        }
      }
    }
    cout << time << "\n";
    s.clear();
  }
  return 0;
}
