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;
map<pair<pair<int, int>, char>, int> M;
int t;
string s;
int main() {
  ios_base::sync_with_stdio(NULL);
  cin.tie(NULL);
  cout.tie(NULL);
  if (fopen(""
            ".inp",
            "r")) {
    freopen(
        ""
        ".inp",
        "r", stdin);
    freopen(
        ""
        ".out",
        "w", stdout);
  }
  cin >> t;
  while (t--) {
    cin >> s;
    int n = s.size();
    s = ' ' + s;
    M.clear();
    int u = 1, v = 1;
    long long ans = 0;
    for (int i = 1; i <= n; ++i) {
      pair<pair<int, int>, char> k = {{u, v}, s[i]};
      if (M[k] == 1)
        ans += 1;
      else
        ans += 5;
      M[k] = 1;
      if (s[i] == 'S') {
        u += 1;
        k = {{u, v}, 'N'};
        M[k] = 1;
      }
      if (s[i] == 'N') {
        u -= 1;
        k = {{u, v}, 'S'};
        M[k] = 1;
      }
      if (s[i] == 'W') {
        v += 1;
        k = {{u, v}, 'E'};
        M[k] = 1;
      }
      if (s[i] == 'E') {
        v -= 1;
        k = {{u, v}, 'W'};
        M[k] = 1;
      }
    }
    cout << ans << '\n';
  }
  return 0;
}
