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;
bool myComparison(const pair<int, int> &a, const pair<int, int> &b) {
  return a.second > b.second;
};
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n, ans = 0;
    string s;
    cin >> s;
    n = s.length();
    int l = -1, r = n;
    map<pair<int, int>, set<char> > vis;
    pair<int, int> cur = {0, 0};
    for (int i = 0; i < n; ++i) {
      if (s[i] == 'W' || s[i] == 'E')
        vis[cur].insert('W' + 'E' - s[i]);
      else
        vis[cur].insert('N' + 'S' - s[i]);
      if (s[i] == 'S') --cur.first;
      if (s[i] == 'N') ++cur.first;
      if (s[i] == 'E') ++cur.second;
      if (s[i] == 'W') --cur.second;
      if (vis.count(cur)) {
        if ((vis[cur].find(s[i]) != vis[cur].end()))
          ans++;
        else
          ans += 5;
      } else
        ans += 5;
      vis[cur].insert(s[i]);
    }
    { cout << ans << endl; }
  }
  return 0;
}
