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 uint = unsigned int;
struct Pos {
  int x = 0, y = 0;
  bool operator<(Pos const &other) const {
    if (x - other.x == 0) {
      return y < other.y;
    } else {
      return x < other.x;
    }
  }
};
int time_ski(std::string const &str) {
  std::set<std::pair<Pos, Pos>> visited;
  Pos current;
  int time = 0;
  for (auto ch : str) {
    auto prev = current;
    switch (ch) {
      case 'N':
        current.y += 1;
        break;
      case 'S':
        current.y -= 1;
        break;
      case 'E':
        current.x += 1;
        break;
      case 'W':
        current.x -= 1;
        break;
      default:
        std::terminate();
    }
    if (visited.find({prev, current}) != visited.end() ||
        visited.find({current, prev}) != visited.end()) {
      time += 1;
    } else {
      time += 5;
      visited.insert({prev, current});
    }
  }
  return time;
}
int main() {
  std::string str;
  std::getline(std::cin, str);
  uint t;
  std::istringstream s{str};
  s >> t;
  for (uint i = 0; i < t; ++i) {
    std::getline(std::cin, str);
    std::cout << time_ski(str) << (i == t - 1 ? "" : "\n");
  }
  return 0;
}
