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;
const int MAXN = 500100, MAXK = 23;
const long long MOD = 998244353;
const long long INF = 1e18;
const long double PI = asin(1) * 2;
void setIO() {
  ios_base::sync_with_stdio(0);
  cin.tie(nullptr);
}
void setIO(string second) {
  setIO();
  freopen((second + ".in").c_str(), "r", stdin);
  freopen((second + ".out").c_str(), "w", stdout);
}
set<pair<pair<int, int>, pair<int, int>>> been;
int main() {
  setIO();
  int t;
  cin >> t;
  while (t--) {
    been.clear();
    string second;
    cin >> second;
    int x = 0, y = 0;
    long long ans = 0;
    for (char c : second) {
      int pX = x, pY = y;
      if (c == 'N')
        y++;
      else if (c == 'S')
        y--;
      else if (c == 'E')
        x++;
      else
        x--;
      if (been.count({{pX, pY}, {x, y}}))
        ans++;
      else
        ans += 5;
      been.insert({{pX, pY}, {x, y}});
      been.insert({{x, y}, {pX, pY}});
    }
    cout << ans << "\n";
  }
  return 0;
}
