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 fast() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
}
vector<string> vec_splitter(string s) {
  s += ',';
  vector<string> res;
  while (!s.empty()) {
    res.push_back(s.substr(0, s.find(',')));
    s = s.substr(s.find(',') + 1);
  }
  return res;
}
void debug_out(vector<string> __attribute__((unused)) args,
               __attribute__((unused)) int idx,
               __attribute__((unused)) int LINE_NUM) {
  cerr << endl;
}
template <typename Head, typename... Tail>
void debug_out(vector<string> args, int idx, int LINE_NUM, Head H, Tail... T) {
  if (idx > 0)
    cerr << ", ";
  else
    cerr << "Line(" << LINE_NUM << ") ";
  stringstream ss;
  ss << H;
  cerr << args[idx] << " = " << ss.str();
  debug_out(args, idx + 1, LINE_NUM, T...);
}
double get_time() { return 1.0 * clock() / CLOCKS_PER_SEC; }
int32_t main() {
  int t;
  cin >> t;
  while (t--) {
    map<pair<pair<int, int>, pair<int, int> >, bool> m;
    string s;
    cin >> s;
    int x = 0, y = 0;
    int res = 0;
    for (char c : s) {
      int px = x, py = y;
      if (c == 'N')
        y += 1;
      else if (c == 'S')
        y -= 1;
      else if (c == 'E')
        x += 1;
      else
        x -= 1;
      if (m[{{px, py}, {x, y}}])
        res++;
      else {
        m[{{px, py}, {x, y}}] = m[{{x, y}, {px, py}}] = 1;
        res += 5;
      }
    }
    cout << res << '\n';
  }
}
