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;
struct trip {
  long long a, b, c;
};
char opp(char ch) {
  if (ch == 'N') return 'S';
  if (ch == 'S') return 'N';
  if (ch == 'E') return 'W';
  if (ch == 'W') return 'E';
}
int main() {
  long long test;
  cin >> test;
  for (int t = 0; t < test; t++) {
    long long max_val = INT_MAX, min_val = -INT_MAX, ans = 0, count_it = 0,
              len = 0;
    bool flag = false;
    string direction;
    int n = 0, e = 0;
    cin >> direction;
    len = direction.size();
    map<pair<pair<long long, long long>, pair<long long, long long> >, int>
        maps;
    long long pren = 0, pree = 0;
    for (int i = 0; i < len; i++) {
      pren = n;
      pree = e;
      if (direction[i] == 'N')
        n++;
      else if (direction[i] == 'S')
        n--;
      else if (direction[i] == 'E')
        e++;
      else
        e--;
      if (maps[{{n, e}, {pren, pree}}] == 0 &&
          maps[{{pren, pree}, {n, e}}] == 0)
        ans += 5;
      else
        ans += 1;
      maps[{{n, e}, {pren, pree}}]++;
      maps[{{pren, pree}, {n, e}}]++;
    }
    cout << ans << "\n";
  }
  return 0;
}
