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>
#pragma GCC optimize("Ofast")
using namespace std;
void manupulate(string &str) {
  for (long long int i = 0; i < str.size(); ++i) str[i] -= 48;
}
bool isprime(long long int n) {
  if (n <= 1) return false;
  for (long long int i = 2; i * i <= n; ++i)
    if (n % i == 0) return false;
  return true;
}
long long int expo(long long int a, long long int n) {
  if (n == 0) return 1;
  if (n == 1) return a;
  long long int tmp = expo(a, n / 2);
  tmp = (tmp * tmp) % (long)(1e9 + 7);
  if (n & 1) tmp = (tmp * a) % (long)(1e9 + 7);
  return tmp;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long int t;
  cin >> t;
  while (t--) {
    string str;
    cin >> str;
    set<array<long long int, 4> > s;
    long long int ans = 0;
    long long int x = 0, y = 0;
    for (long long int i = 0; i < str.size(); ++i) {
      long long int x1 = x, y1 = y;
      if (str[i] == 'N')
        ++y1;
      else if (str[i] == 'E')
        ++x1;
      else if (str[i] == 'S')
        --y1;
      else
        --x1;
      array<long long int, 4> tmp = {x, y, x1, y1};
      array<long long int, 4> tmp2 = {x1, y1, x, y};
      if (s.find(tmp) != s.end() || s.find(tmp2) != s.end())
        ++ans;
      else {
        s.insert(tmp);
        s.insert(tmp2);
        ans += 5;
      }
      x = x1, y = y1;
    }
    cout << ans << "\n";
  }
  return 0;
}
