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;
class DebugStream {
} LOG;
template <typename T>
DebugStream &operator<<(DebugStream &s, const T &) {
  return s;
}
const int MAX = 100100;
char s[MAX];
void apply(int &x, int &y, char c) {
  if (c == 'N') y++;
  if (c == 'S') y--;
  if (c == 'E') x++;
  if (c == 'W') x--;
}
int main() {
  int T;
  scanf("%d", &T);
  for (int t = 0; t < int(T); t++) {
    scanf("%s", s);
    int len = strlen(s);
    int x = 0, y = 0;
    set<pair<pair<int, int>, pair<int, int> > > visited;
    long long ans = 0;
    for (int i = 0; i < int(len); i++) {
      int nx = x, ny = y;
      apply(nx, ny, s[i]);
      pair<int, int> a(x, y), b(nx, ny);
      if (a > b) swap(a, b);
      if (visited.count(make_pair(a, b)) == 1)
        ans += 1;
      else
        ans += 5;
      visited.insert(make_pair(a, b));
      x = nx;
      y = ny;
    }
    printf("%lld\n", ans);
  }
}
