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;
long long int arr[105];
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int n, x, y, t, l, m, k = 0, r = 0, a, b, c, d, p = 0, q = 0, i, j,
                                  ans = 0, ans1 = 0;
  string s1, s;
  char ch;
  cin >> t;
  map<pair<pair<long long int, long long int>,
           pair<long long int, long long int> >,
      bool>
      M;
  while (t--) {
    cin >> s;
    long long int x1, y1;
    x = y = 0;
    ans = 0;
    for (i = 0; i <= (long long int)s.size() - 1; i++) {
      x1 = x;
      y1 = y;
      if (s[i] == 'N') y++;
      if (s[i] == 'S') y--;
      if (s[i] == 'E') x++;
      if (s[i] == 'W') x--;
      if (M[make_pair(make_pair(x, y), make_pair(x1, y1))] == 1)
        ans += 1;
      else
        ans += 5;
      M[make_pair(make_pair(x, y), make_pair(x1, y1))] = 1;
      M[make_pair(make_pair(x1, y1), make_pair(x, y))] = 1;
    }
    cout << ans << "\n";
    M.clear();
  }
  return 0;
}
