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 vout(vector<long long> ans) {
  for (auto ele : ans) cout << ele << " ";
  cout << endl;
}
vector<long long> adj[200001];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cout.precision(8);
  cout << fixed;
  long long t;
  cin >> t;
  while (t--) {
    map<pair<pair<long long, long long>, pair<long long, long long>>, bool> vis;
    string s;
    cin >> s;
    long long x = 0;
    long long ans = 0;
    long long y = 0;
    vis[make_pair(make_pair(x, y), make_pair(x, y))] = true;
    for (auto ele : s) {
      long long ox = x;
      long long oy = y;
      if (ele == 'N') {
        y += 1;
      } else if (ele == 'S') {
        y -= 1;
      } else if (ele == 'W') {
        x -= 1;
      } else {
        x += 1;
      }
      if (vis[make_pair(make_pair(ox, oy), make_pair(x, y))])
        ans += 1;
      else {
        vis[make_pair(make_pair(ox, oy), make_pair(x, y))] = true;
        vis[make_pair(make_pair(x, y), make_pair(ox, oy))] = true;
        ans += 5;
      }
    }
    cout << ans << endl;
  }
  return 0;
}
