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(void) {
  int n;
  cin >> n;
  while (n--) {
    string s;
    cin >> s;
    int ans = 0;
    map<pair<int, int>, set<char> > m;
    pair<int, int> pos = {0, 0};
    for (auto e : s) {
      if (m[pos].find(e) != m[pos].end()) {
        if (e == 'N') pos.second++;
        if (e == 'S') pos.second--;
        if (e == 'E') pos.first++;
        if (e == 'W') pos.first--;
        ans += 1;
        continue;
      }
      m[pos].insert(e);
      if (e == 'N') {
        pos.second++;
        int vis = 0;
        for (auto k : m[pos]) {
          if (k == 'S') {
            ans += 1;
            vis = 1;
            break;
          }
        }
        if (!vis) ans += 5;
      } else if (e == 'S') {
        pos.second--;
        int vis = 0;
        for (auto k : m[pos]) {
          if (k == 'N') {
            ans += 1;
            vis = 1;
            break;
          }
        }
        if (!vis) ans += 5;
      } else if (e == 'W') {
        pos.first--;
        int vis = 0;
        for (auto k : m[pos])
          if (k == 'E') {
            ans += 1;
            vis = 1;
            break;
          }
        if (!vis) ans += 5;
      } else if (e == 'E') {
        pos.first++;
        int vis = 0;
        for (auto k : m[pos])
          if (k == 'W') {
            ans += 1;
            vis = 1;
            break;
          }
        if (!vis) ans += 5;
      }
    }
    cout << ans << '\n';
  }
  return (0);
}
