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;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  int t;
  cin >> t;
  while (t-- > 0) {
    string s;
    set<pair<pair<long long, long long>, pair<long long, long long> > > st;
    cin >> s;
    pair<long long, long long> curr = {0, 0};
    long long ans = 0;
    for (int i = 0; i < s.length(); i++) {
      if (s[i] == 'N') {
        pair<long long, long long> curr2 = curr;
        curr = {curr.first, curr.second + 1};
        if (st.find({curr2, curr}) == st.end()) {
          st.insert({curr2, curr});
          st.insert({curr, curr2});
          ans += 5;
        } else {
          ans += 1;
        }
      }
      if (s[i] == 'S') {
        pair<long long, long long> curr2 = curr;
        curr = {curr.first, curr.second - 1};
        if (st.find({curr2, curr}) == st.end()) {
          st.insert({curr2, curr});
          st.insert({curr, curr2});
          ans += 5;
        } else {
          ans += 1;
        }
      }
      if (s[i] == 'E') {
        pair<long long, long long> curr2 = curr;
        curr = {curr.first - 1, curr.second};
        if (st.find({curr2, curr}) == st.end()) {
          st.insert({curr2, curr});
          st.insert({curr, curr2});
          ans += 5;
        } else {
          ans += 1;
        }
      }
      if (s[i] == 'W') {
        pair<long long, long long> curr2 = curr;
        curr = {curr.first + 1, curr.second};
        if (st.find({curr2, curr}) == st.end()) {
          st.insert({curr2, curr});
          st.insert({curr, curr2});
          ans += 5;
        } else {
          ans += 1;
        }
      }
    }
    cout << ans << "\n";
  }
}
