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;
vector<int> v;
vector<vector<int>> ed;
void dfs(int i) {
  v[i] = 1;
  for (auto& j : ed[i])
    if (!v[j]) dfs(j);
}
void Fast_IO() {
  ios::sync_with_stdio(0);
  cin.tie(0);
}
int main() {
  Fast_IO();
  int t;
  cin >> t;
  while (t--) {
    string st;
    cin >> st;
    int m[] = {0, 0, 1, -1};
    int n[] = {1, -1, 0, 0};
    int x = 0;
    int y = 0;
    int ans = 0;
    set<pair<pair<int, int>, pair<int, int>>> second;
    for (int i = 0; i < st.size(); i++) {
      int d;
      if (st[i] == 'N') d = 0;
      if (st[i] == 'S') d = 1;
      if (st[i] == 'E') d = 2;
      if (st[i] == 'W') d = 3;
      if ((second.find({{x, y}, {x + m[d], y + n[d]}})) != second.end() ||
          (second.find({{x + m[d], y + n[d]}, {x, y}}) != second.end()))
        ans++;
      else {
        ans += 5;
        second.insert({{x, y}, {x + m[d], y + n[d]}});
      }
      x = x + m[d];
      y = y + n[d];
    }
    cout << ans << '\n';
  }
  return 0;
}
