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;
void solve() {
  map<pair<pair<int, int>, pair<int, int> >, int> visto;
  string stringa;
  cin >> stringa;
  int x = 0, y = 0;
  pair<int, int> punto;
  long long int risultato = 0;
  for (int a = 0; a < stringa.size(); a++) {
    if (stringa[a] == 'N') {
      if (!visto.count({{x, y}, {x + 1, y}}) &&
          !visto.count({{x + 1, y}, {x, y}})) {
        visto[{{x, y}, {x + 1, y}}]++;
        visto[{{x + 1, y}, {x, y}}]++;
        risultato += 5;
      } else
        risultato++;
      x++;
    } else if (stringa[a] == 'S') {
      if (!visto.count({{x, y}, {x - 1, y}}) &&
          !visto.count({{x - 1, y}, {x, y}})) {
        visto[{{x, y}, {x - 1, y}}]++;
        visto[{{x - 1, y}, {x, y}}]++;
        risultato += 5;
      } else
        risultato++;
      x--;
    } else if (stringa[a] == 'W') {
      if (!visto.count({{x, y}, {x, y - 1}}) &&
          !visto.count({{x, y - 1}, {x, y}})) {
        visto[{{x, y}, {x, y - 1}}]++;
        visto[{{x, y - 1}, {x, y}}]++;
        risultato += 5;
      } else
        risultato++;
      y--;
    } else {
      if (!visto.count({{x, y}, {x, y + 1}}) &&
          !visto.count({{x, y + 1}, {x, y}})) {
        visto[{{x, y}, {x, y + 1}}]++;
        visto[{{x, y + 1}, {x, y}}]++;
        risultato += 5;
      } else
        risultato++;
      y++;
    }
  }
  cout << risultato << endl;
}
int main() {
  int test;
  cin >> test;
  while (test--) solve();
}
