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("O3")
using namespace std;
long long ppow(long long x, long long y, long long mod) {
  long long res = 1;
  while (y) {
    if (y & 1) res = res * x % mod;
    y >>= 1;
    x = x * x % mod;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    string second;
    cin >> second;
    map<pair<int, int>, map<pair<int, int>, bool>> was;
    long long ans = 5;
    int x = 0, y = 0;
    int xx = 0, yy = 0;
    char c = second[0];
    if (c == 'S') yy--;
    if (c == 'N') yy++;
    if (c == 'W') xx--;
    if (c == 'E') xx++;
    was[{x, y}][{xx, yy}] = was[{xx, yy}][{x, y}] = 1;
    x = xx, y = yy;
    for (int i = 1; i < second.size(); ++i) {
      c = second[i];
      xx = x, yy = y;
      if (c == 'S') yy--;
      if (c == 'N') yy++;
      if (c == 'W') xx--;
      if (c == 'E') xx++;
      if (was[{x, y}][{xx, yy}])
        ans++;
      else
        ans += 5;
      was[{x, y}][{xx, yy}] = was[{xx, yy}][{x, y}] = 1;
      x = xx, y = yy;
    }
    cout << ans << "\n";
  }
}
