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 dx8[] = {0, 0, 1, 1, 1, -1, -1, -1};
int dy8[] = {1, -1, 1, -1, 0, 0, -1, 1};
int dx4[] = {0, 0, 1, -1};
int dy4[] = {1, -1, 0, 0};
const double pi = acos(-1.0);
const double eps = 1e-6;
const int mod = (int)1e9 + 7;
const int maxn = 300005;
const int logn = 20;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  cin >> t;
  while (t--) {
    string s;
    cin >> s;
    pair<int, int> pre = make_pair(0, 0);
    set<pair<pair<int, int>, pair<int, int>>> st;
    int n = s.size();
    long long tot = 0LL;
    for (int i = 0; i < (n); i++) {
      pair<int, int> now = pre;
      if (s[i] == 'N')
        now.second++;
      else if (s[i] == 'S')
        now.second--;
      else if (s[i] == 'E')
        now.first++;
      else
        now.first--;
      if (st.count(make_pair(pre, now)) > 0 ||
          st.count(make_pair(now, pre)) > 0)
        tot++;
      else
        tot += 5;
      st.insert(make_pair(pre, now));
      st.insert(make_pair(now, pre));
      pre = now;
    }
    cout << tot << "\n";
  }
  return 0;
}
