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;
set<pair<pair<int, int>, char> > se;
int main() {
  int t;
  cin >> t;
  while (t--) {
    string s;
    long long int now = 0;
    int a, b;
    a = b = 0;
    cin >> s;
    se.clear();
    for (auto it : s) {
      switch (it) {
        case 'N': {
          a++;
          if (se.count({{a, b}, 'S'}))
            now++;
          else {
            now += 5;
            se.insert({{a, b}, 'S'});
            se.insert({{a - 1, b}, 'N'});
          }
          break;
        }
        case 'E': {
          b--;
          if (se.count({{a, b}, 'W'}))
            now++;
          else {
            now += 5;
            se.insert({{a, b}, 'W'});
            se.insert({{a, b + 1}, 'E'});
          }
          break;
        }
        case 'W': {
          b++;
          if (se.count({{a, b}, 'E'}))
            now++;
          else {
            now += 5;
            se.insert({{a, b}, 'E'});
            se.insert({{a, b - 1}, 'W'});
          }
          break;
        }
        case 'S': {
          a--;
          if (se.count({{a, b}, 'N'}))
            now++;
          else {
            now += 5;
            se.insert({{a, b}, 'N'});
            se.insert({{a + 1, b}, 'S'});
          }
          break;
        }
      }
    }
    cout << now << endl;
  }
}
