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 main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    string s;
    cin >> s;
    map<pair<pair<int, int>, pair<int, int>>, bool> mp;
    int x = 0, y = 0, cnt = 0;
    for (int i = 0; i < s.size(); ++i) {
      if (s[i] == 'W') {
        if (mp[{{x, y}, {x, y - 1}}]) {
          ++cnt;
        } else {
          cnt += 5;
          mp[{{x, y}, {x, y - 1}}] = true;
        }
        --y;
      } else if (s[i] == 'E') {
        if (mp[{{x, y + 1}, {x, y}}]) {
          ++cnt;
        } else {
          cnt += 5;
          mp[{{x, y + 1}, {x, y}}] = true;
        }
        ++y;
      } else if (s[i] == 'N') {
        if (mp[{{x, y}, {x + 1, y}}]) {
          ++cnt;
        } else {
          cnt += 5;
          mp[{{x, y}, {x + 1, y}}] = true;
        }
        ++x;
      } else {
        if (mp[{{x - 1, y}, {x, y}}]) {
          ++cnt;
        } else {
          cnt += 5;
          mp[{{x - 1, y}, {x, y}}] = true;
        }
        --x;
      }
    }
    cout << cnt << "\n";
  }
  return 0;
}
