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>
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops")
#pragma GCC optimize("no-stack-protector,fast-math")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
template <class S, class T>
inline S smax(S &l, T r) {
  return l = max(l, (S)r);
}
template <class S, class T>
inline bool smin(S &l, T r) {
  return l = min(l, (S)r);
}
const int MAXN = 1e5 + 10;
set<pair<pair<int, int>, pair<int, int>>> st;
void solve() {
  string s;
  cin >> s;
  st.clear();
  int x = 0, y = 0;
  int n = s.size();
  long long ans = 0;
  for (int i = 0; i < n; ++i) {
    int x_ = x, y_ = y;
    if (s[i] == 'N') x_++;
    if (s[i] == 'S') x_--;
    if (s[i] == 'W') y_--;
    if (s[i] == 'E') y_++;
    int osz = st.size();
    st.insert({{x, y}, {x_, y_}});
    st.insert({{x_, y_}, {x, y}});
    int nsz = st.size();
    if (osz == nsz)
      ans++;
    else
      ans += 5;
    x = x_;
    y = y_;
  }
  cout << ans << '\n';
}
int32_t main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}
