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;
int dir[4][2] = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
const long long INF = 0x7f7f7f7f7f7f7f7f;
const int inf = 0x3f3f3f3f;
const double pi = 3.14159265358979323846;
const int mod = 998244353;
const int N = 2e5 + 5;
inline long long read() {
  long long x = 0;
  bool f = true;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = false;
    c = getchar();
  }
  while (c >= '0' && c <= '9')
    x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
  return f ? x : -x;
}
long long gcd(long long m, long long n) { return n == 0 ? m : gcd(n, m % n); }
long long lcm(long long m, long long n) { return m * n / gcd(m, n); }
int main() {
  int T;
  cin >> T;
  while (T--) {
    queue<char> q;
    string s;
    cin >> s;
    int res = 0;
    map<pair<int, int>, set<pair<int, int>>> mp;
    pair<int, int> cur = {0, 0};
    for (int i = 0; i < s.size(); i++) {
      pair<int, int> t = cur;
      if (s[i] == 'N')
        cur.second++;
      else if (s[i] == 'S')
        cur.second--;
      else if (s[i] == 'E')
        cur.first++;
      else if (s[i] == 'W')
        cur.first--;
      if (mp[t].count(cur)) {
        res++;
      } else {
        mp[t].insert(cur);
        mp[cur].insert(t);
        res += 5;
      }
    }
    cout << res << endl;
  }
  return 0;
}
