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;
bool isPrime(long long n) {
  if (n < 2) return false;
  for (long long x = 2; x * x <= n; x++) {
    if (n % x == 0) return false;
  }
  return true;
}
bool isPerfectSquare(long double x) {
  long double sr = sqrt(x);
  return ((sr - floor(sr)) == 0);
}
int gcd(int a, int b) {
  if (a == 0) return b;
  if (b == 0) return a;
  if (a == b) return a;
  if (a > b) return gcd(a - b, b);
  return gcd(a, b - a);
}
int32_t main() {
  int t;
  cin >> t;
  while (t--) {
    string s;
    cin >> s;
    int n = (int)s.size();
    int ans = 0;
    map<pair<pair<int, int>, pair<int, int>>, int> m;
    int x = 0, y = 0;
    for (int i = 0; i < n; i++) {
      int nowx = x, nowy = y;
      if (s[i] == 'N') y++;
      if (s[i] == 'S') y--;
      if (s[i] == 'E') x--;
      if (s[i] == 'W') x++;
      if (m[{{x, y}, {nowx, nowy}}] || m[{{nowx, nowy}, {x, y}}]) {
        ans++;
      } else {
        ans += 5;
      }
      m[{{x, y}, {nowx, nowy}}]++;
    }
    cout << ans << "\n";
  }
}
