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 Int = long long;
int PREP = (cin.tie(nullptr), ios::sync_with_stdio(false),
            cout << fixed << setprecision(9), 0);
int solve(const string &S) {
  set<Int> se;
  Int x = 1e5, y = 1e5;
  int ans = 0;
  for (char c : S) {
    Int nx = 0, ny = 0;
    if (c == 'S') nx = x, ny = y - 1;
    if (c == 'N') nx = x, ny = y + 1;
    if (c == 'W') nx = x - 1, ny = y;
    if (c == 'E') nx = x + 1, ny = y;
    bool b = false;
    Int id = 0;
    if (nx == x)
      id = 0 + x * (Int)1e9 + min(y, ny);
    else
      id = (Int)1e18 + min(x, nx) * (Int)1e9 + y;
    if (se.count(id))
      b = 1;
    else
      se.insert(id);
    ans += (b ? 1 : 5);
    x = nx, y = ny;
  }
  return ans;
}
int main() {
  int T;
  cin >> T;
  while (T--) {
    string S;
    cin >> S;
    cout << solve(S) << '\n';
  }
  return 0;
}
