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 compare(const pair<long long, long long>& a,
             const pair<long long, long long>& b) {
  if (a.first == b.first) return a.second < b.second;
  return a.first > b.first;
}
bool sortbysec(const pair<long long, long long>& a,
               const pair<long long, long long>& b) {
  return a.second < b.second;
}
signed main() {
  long long t;
  cin >> t;
  while (t--) {
    string s;
    cin >> s;
    map<pair<pair<long long, long long>, pair<long long, long long> >,
        long long>
        m;
    long long time = 0;
    long long x = 0, y = 0;
    for (auto i : s) {
      long long x1 = x, y1 = y;
      if (i == 'N') y++;
      if (i == 'S') y--;
      if (i == 'E') x++;
      if (i == 'W') x--;
      pair<pair<long long, long long>, pair<long long, long long> > p;
      if (x == x1)
        p = {{x, min(y1, y)}, {x, max(y1, y)}};
      else
        p = {{min(x, x1), y}, {max(x, x1), y}};
      if (m[p] == 1)
        time++;
      else
        time += 5;
      m[p] = 1;
    }
    cout << time << endl;
  }
}
