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;
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int t;
  cin >> t;
  while (t--) {
    long long int x = 0, y = 0;
    string str;
    cin >> str;
    long long int answer = 0;
    set<pair<pair<long long int, long long int>,
             pair<long long int, long long int>>>
        st;
    long long int j;
    long long int n = str.size();
    for (j = 0; j < n; j++) {
      long long int val1 = x, val2 = y;
      if (str[j] == 'N') y++;
      if (str[j] == 'S') y--;
      if (str[j] == 'E') x++;
      if (str[j] == 'W') x--;
      pair<long long int, long long int> p1, p2;
      p1.first = x;
      p1.second = y;
      p2.first = val1;
      p2.second = val2;
      auto it = st.find({p1, p2});
      if (it == st.end()) {
        answer += 5;
        st.insert({p1, p2});
        st.insert({p2, p1});
      } else {
        answer++;
      }
    }
    cout << answer << endl;
  }
}
