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;
void solve() {
  int x = 0, y = 0;
  set<pair<int, int> > x1, y1;
  string s;
  cin >> s;
  for (int i = 0; i < s.size(); i++) {
    if (s[i] == 'N') {
      y++;
      x1.insert(make_pair(x, y));
    } else if (s[i] == 'S') {
      y--;
      x1.insert(make_pair(x, y + 1));
    } else if (s[i] == 'W') {
      x++;
      y1.insert(make_pair(x, y));
    } else if (s[i] == 'E') {
      x--;
      y1.insert(make_pair(x + 1, y));
    }
  }
  int ans = x1.size() + y1.size();
  cout << ans * 5 + (s.size() - ans) << endl;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}
