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 test;
string str;
set<pair<pair<int, int>, pair<int, int>>> s;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> test;
  while (test--) {
    s.clear();
    cin >> str;
    int x = 0, y = 0, u, v;
    int ans = 0;
    for (char ch : str) {
      u = x, v = y;
      if (ch == 'N') {
        v++;
      }
      if (ch == 'E') {
        u++;
      }
      if (ch == 'S') {
        v--;
      }
      if (ch == 'W') {
        u--;
      }
      if (s.find(make_pair(make_pair(x, y), make_pair(u, v))) == s.end() &&
          s.find(make_pair(make_pair(u, v), make_pair(x, y))) == s.end()) {
        ans += 5;
        s.insert(make_pair(make_pair(x, y), make_pair(u, v)));
      } else {
        ans++;
      }
      x = u, y = v;
    }
    cout << ans << '\n';
  }
  return 0;
}
