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 modularPower(long long int x, unsigned long long int y,
                           long long int mod) {
  long long int res = 1;
  x %= mod;
  while (y > 0) {
    if (y & 1) res = (res * x) % mod;
    y >>= 1;
    x = (x * x) % mod;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cerr.tie(NULL);
  cout.tie(NULL);
  ;
  int tests = 1;
  cin >> tests;
  while (tests--) {
    string s;
    cin >> s;
    int ans = 0;
    int x = 0, y = 0;
    set<tuple<int, int, int, int>> moves;
    for (auto ch : s) {
      int foo = x;
      int bar = y;
      if (ch == 'N') y++;
      if (ch == 'S') y--;
      if (ch == 'W') x--;
      if (ch == 'E') x++;
      if (moves.find({x, y, foo, bar}) == moves.end() and
          moves.find({foo, bar, x, y}) == moves.end()) {
        ans += 5;
        moves.insert({x, y, foo, bar});
      } else {
        ans += 1;
      }
    }
    cout << ans << "\n";
  }
  return 0;
}
