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;
const long long logN = 20;
const long long M = 1000000007;
const long long MM = 998244353;
const long long INF = 1e12;
const long long N = 100005;
bool sortb(pair<long long, long long> const &a,
           pair<long long, long long> const &b) {
  return a.second < b.second;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t, i;
  cin >> t;
  while (t--) {
    string s;
    cin >> s;
    long long h = 0, v = 0;
    set<pair<pair<long long, long long>, pair<long long, long long> > > st;
    long long ans = 0, h0 = 0, v0 = 0;
    for (i = 0; i < s.size(); i++) {
      if (s[i] == 'E') {
        h++;
      } else if (s[i] == 'W') {
        h--;
      } else if (s[i] == 'N') {
        v++;
      } else {
        v--;
      }
      if (i == 0) {
        st.insert(make_pair(make_pair(h, v), make_pair(0, 0)));
        ans = ans + 5;
      } else {
        if (st.find(make_pair(make_pair(h, v), make_pair(h0, v0))) !=
                st.end() ||
            st.find(make_pair(make_pair(h0, v0), make_pair(h, v))) !=
                st.end()) {
          ans = ans + 1;
        } else {
          st.insert(make_pair(make_pair(h, v), make_pair(h0, v0)));
          ans = ans + 5;
        }
      }
      h0 = h;
      v0 = v;
    }
    cout << ans << endl;
  }
  return 0;
}
