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;
using namespace std;
const long long int M = 1e9 + 7;
void solve() {
  string s;
  cin >> s;
  long long int x = 0, y = 0;
  set<pair<pair<int, int>, pair<int, int>>> ss;
  long long int count = 0;
  for (int i = 0; i < s.length(); i++) {
    int px = x, py = y;
    if (s[i] == 'N') {
      y += 1;
    } else if (s[i] == 'S') {
      y -= 1;
    } else if (s[i] == 'W') {
      x -= 1;
    } else if (s[i] == 'E') {
      x += 1;
    }
    pair<int, int> p = {x, y};
    pair<int, int> p1 = {px, py};
    if (ss.find({p, p1}) != ss.end() || ss.find({p1, p}) != ss.end()) {
      count++;
    } else {
      count += 5;
    }
    ss.insert({p, p1});
    ss.insert({p1, p});
  }
  cout << count << endl;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  srand(time(NULL));
  long long int t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}
