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;
using ll = long long;
using ld = double;
void solve() {
  string s;
  cin >> s;
  ll n = s.length();
  ll x = 0, y = 0, k = 0, z = 0;
  set<pair<pair<ll, ll>, pair<ll, ll>>> st;
  for (ll i = 0; i < n; i++) {
    if (s[i] == 'N') {
      if (!st.count({{x, y}, {x, y + 1}})) {
        z += 5;
      } else
        z += 1;
      st.insert({{x, y}, {x, y + 1}});
      st.insert({{x, y + 1}, {x, y}});
      y++;
    }
    if (s[i] == 'E') {
      if (!st.count({{x, y}, {x + 1, y}})) {
        z += 5;
      } else
        z += 1;
      st.insert({{x + 1, y}, {x, y}});
      st.insert({{x, y}, {x + 1, y}});
      x++;
    }
    if (s[i] == 'S') {
      if (!st.count({{x, y}, {x, y - 1}})) {
        z += 5;
      } else
        z += 1;
      st.insert({{x, y}, {x, y - 1}});
      st.insert({{x, y - 1}, {x, y}});
      y--;
    }
    if (s[i] == 'W') {
      if (!st.count({{x, y}, {x - 1, y}})) {
        z += 5;
      } else
        z += 1;
      st.insert({{x, y}, {x - 1, y}});
      st.insert({{x - 1, y}, {x, y}});
      x--;
    }
  }
  cout << z << '\n';
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ll t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
}
