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() {
  string s;
  cin >> s;
  set<pair<pair<long long, long long>, pair<long long, long long>>> v;
  long long a = 0;
  long long b = 0;
  long long c = 0;
  long long d = 0;
  long long sum = 0;
  long long n = int((s).size());
  for (long long i = 0; i < n; i++) {
    if (s[i] == 'N') c = a + 1;
    if (s[i] == 'S') c = a - 1;
    if (s[i] == 'W') d = b - 1;
    if (s[i] == 'E') d = b + 1;
    if (v.find(make_pair(make_pair(a, b), make_pair(c, d))) == v.end()) {
      sum += 5;
      v.insert(make_pair(make_pair(a, b), make_pair(c, d)));
      v.insert(make_pair(make_pair(c, d), make_pair(a, b)));
    } else
      sum += 1;
    a = c;
    b = d;
  }
  cout << sum;
  cout << "\n";
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long test;
  cin >> test;
  while (test--) {
    solve();
  }
  return 0;
}
