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 int modpow(long long int a, long long int b,
                     long long int m = 1000000007) {
  a %= m;
  long long int res = 1;
  while (b) {
    if (b & 1) res = (res * a) % m;
    a = (a * a) % m;
    b >>= 1;
  }
  return res;
}
long long int bpow(long long int a, long long int b) {
  long long int res = 1;
  while (b) {
    if (b & 1) res = res * a;
    a = a * a;
    b >>= 1;
  }
  return res;
}
long long int modinv(long long int a, long long int m = 1000000007) {
  return modpow(a, m - 2, m);
}
void graph(vector<vector<int> > &adj, int m) {
  int x, y;
  for (int i = 0; i < m; i++) {
    cin >> x >> y;
    adj[--x].push_back(--y);
    adj[y].push_back(x);
  }
}
void solve() {
  string str;
  cin >> str;
  set<pair<pair<int, int>, pair<int, int> > > s;
  int currx = 0, curry = 0;
  int time = 0;
  for (auto c : str) {
    int l1 = s.size();
    if (c == 'N') {
      s.insert(make_pair(make_pair(currx, curry), make_pair(currx, curry + 1)));
      s.insert(make_pair(make_pair(currx, curry + 1), make_pair(currx, curry)));
      curry++;
    } else if (c == 'S') {
      s.insert(make_pair(make_pair(currx, curry), make_pair(currx, curry - 1)));
      s.insert(make_pair(make_pair(currx, curry - 1), make_pair(currx, curry)));
      curry--;
    } else if (c == 'E') {
      s.insert(make_pair(make_pair(currx, curry), make_pair(currx + 1, curry)));
      s.insert(make_pair(make_pair(currx + 1, curry), make_pair(currx, curry)));
      currx++;
    } else if (c == 'W') {
      s.insert(make_pair(make_pair(currx, curry), make_pair(currx - 1, curry)));
      s.insert(make_pair(make_pair(currx - 1, curry), make_pair(currx, curry)));
      currx--;
    }
    if (s.size() == l1)
      time += 1;
    else
      time += 5;
  }
  cout << time << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  cin >> t;
  while (t--) solve();
}
