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;
void solve() {
  string moves;
  cin >> moves;
  set<pair<pair<int, int>, pair<int, int>>> visited;
  pair<int, int> start(0, 0);
  pair<int, int> end;
  int ans = 0;
  for (unsigned long int j = 0; j < moves.length(); j++) {
    char i = moves[j];
    if (i == 'N') {
      end.first = start.first;
      end.second = start.second + 1;
    } else if (i == 'S') {
      end.first = start.first;
      end.second = start.second - 1;
    } else if (i == 'E') {
      end.first = start.first + 1;
      end.second = start.second;
    } else {
      end.first = start.first - 1;
      end.second = start.second;
    }
    pair<pair<int, int>, pair<int, int>> path1(start, end);
    pair<pair<int, int>, pair<int, int>> path2(end, start);
    if (visited.count(path1) || visited.count(path2)) {
      ans += 1;
    } else {
      ans += 5;
      visited.insert(path1);
    }
    start = end;
  }
  cout << ans << "\n";
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}
