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 f(long long a) {
  long long cnt = 0, i;
  if (a == 2) return 1;
  for (i = 2; i * i < a; i++) {
    if (a % i == 0) cnt += 2;
  }
  if (i * i == a) cnt++;
  return cnt;
}
vector<long long> a(1001);
long long powM(long long a, long long b, long long m) {
  long long ans = 1;
  while (b > 0) {
    if (b & 1) ans = ans * a % m;
    a = a * a % m;
    b >>= 1;
  }
  return ans % m;
}
long long g(long long a, long long n) {
  if (a < 0)
    return ((a % n) + n) % n;
  else
    return a % n;
}
int main() {
  long long t;
  cin >> t;
  while (t--) {
    string s;
    cin >> s;
    long long x = 0, y = 0;
    long long n = s.length();
    long long ans = 0;
    set<tuple<long long, long long, long long, long long>> st;
    for (int i = 0; i < n; i++) {
      long long xp = x;
      long long yp = y;
      if (s[i] == 'N')
        y++;
      else if (s[i] == 'S')
        y--;
      else if (s[i] == 'E')
        x++;
      else
        x--;
      if (st.find({x, xp, y, yp}) != st.end() &&
          st.find({xp, x, yp, y}) != st.end())
        ans++;
      else {
        st.insert({x, xp, y, yp});
        st.insert({xp, x, yp, y});
        ans += 5;
      }
    }
    cout << ans << "\n";
  }
  return 0;
}
