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;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    string str;
    cin >> str;
    set<pair<pair<int, int>, pair<int, int>>> S;
    auto pre = make_pair(0, 0);
    int ans = 0;
    for (auto ch : str) {
      pair<int, int> next = pre;
      if (ch == 'N') {
        next.first++;
      } else if (ch == 'S') {
        next.first--;
      } else if (ch == 'E') {
        next.second++;
      } else {
        next.second--;
      }
      pair<pair<int, int>, pair<int, int>> d1 = make_pair(pre, next),
                                           d2 = make_pair(next, pre);
      if (S.find(d1) == S.end() && S.find(d2) == S.end()) {
        ans += 5;
        S.insert(d1);
      } else {
        ans += 1;
      }
      pre = next;
    }
    cout << ans << "\n";
  }
  return 0;
}
