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;
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
void solve() {
  string s;
  cin >> s;
  set<pair<long long, long long> > box;
  long long x = 0;
  long long y = 0;
  long long i = 0;
  long long n = s.size();
  long long ans = 0;
  while (i < n) {
    if (s[i] == 'N') {
      y = y + 2;
      if (box.find({x, y - 1}) == box.end()) {
        box.insert({x, y - 1});
        ans = ans + 5;
      } else {
        ans++;
      }
    } else if (s[i] == 'S') {
      y = y - 2;
      if (box.find({x, y + 1}) == box.end()) {
        box.insert({x, y + 1});
        ans = ans + 5;
      } else {
        ans++;
      }
    } else if (s[i] == 'E') {
      x = x + 2;
      if (box.find({x - 1, y}) == box.end()) {
        box.insert({x - 1, y});
        ans = ans + 5;
      } else {
        ans++;
      }
    } else {
      x = x - 2;
      if (box.find({x + 1, y}) == box.end()) {
        box.insert({x + 1, y});
        ans = ans + 5;
      } else {
        ans++;
      }
    }
    i++;
  }
  cout << ans << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout << setprecision(11);
  long long t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}
