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;
const long long int MOD = 1e9 + 7;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  ;
  while (t--) {
    string s;
    cin >> s;
    int x = 0, y = 0;
    long long int ans = 0;
    set<pair<int, int> > map;
    for (auto c : s) {
      if (c == 'N') {
        if (map.find({x, y + 1}) == map.end()) {
          map.insert({x, y + 1});
          ans += 5;
        } else
          ans++;
        y += 2;
      } else if (c == 'S') {
        if (map.find({x, y - 1}) == map.end()) {
          map.insert({x, y - 1});
          ans += 5;
        } else
          ans++;
        y -= 2;
      } else if (c == 'W') {
        if (map.find({x - 1, y}) == map.end()) {
          map.insert({x - 1, y});
          ans += 5;
        } else
          ans++;
        x -= 2;
      } else {
        if (map.find({x + 1, y}) == map.end()) {
          map.insert({x + 1, y});
          ans += 5;
        } else
          ans++;
        x += 2;
      }
    }
    cout << ans << '\n';
  }
  return 0;
}
