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;
vector<long long int> v;
long long int fep(long long int x, long long int y, long long int mod) {
  long long int res = 1;
  while (y > 0) {
    if (y & 1) res = (res * x) % mod;
    y = y >> 1;
    x = (x * x) % mod;
  }
  return res % mod;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int n, i, m, j, k, x, y, a, b, t;
  string s;
  cin >> t;
  while (t--) {
    set<pair<pair<long long int, long long int>,
             pair<long long int, long long int> > >
        st;
    cin >> s;
    n = s.size();
    a = 0;
    long long int px = 0, py = 0;
    x = 0, y = 0;
    m = 1000000;
    for (i = 0; i < n; i++) {
      if (s[i] == 'S') y--;
      if (s[i] == 'N') y++;
      if (s[i] == 'E') x++;
      if (s[i] == 'W') x--;
      if (st.find({{x, y}, {px, py}}) != st.end() ||
          st.find({{px, py}, {x, y}}) != st.end())
        a++;
      else {
        a += 5;
        st.insert({{x, y}, {px, py}});
      }
      px = x, py = y;
    }
    cout << a << endl;
  }
  return 0;
}
