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 ll = long long;
constexpr int INF = 1073741823;
constexpr int MINF = -1073741823;
constexpr ll LINF = ll(4661686018427387903);
constexpr ll MOD = 1000000007;
constexpr long double eps = 1e-9;
const long double PI = acosl(-1.0L);
using namespace std;
void scans(string& str) {
  char c;
  str = "";
  scanf("%c", &c);
  if (c == '\n') scanf("%c", &c);
  while (c != '\n' && c != -1 && c != ' ') {
    str += c;
    scanf("%c", &c);
  }
}
void scanc(char& str) {
  char c;
  scanf("%c", &c);
  if (c == -1) return;
  while (c == '\n') {
    scanf("%c", &c);
  }
  str = c;
}
double acot(double x) { return PI / 2 - atan(x); }
ll LSB(ll n) { return (n & (-n)); }
int main() {
  int t;
  scanf("%d", &t);
  int cnt = 1;
  while (t--) {
    string s;
    scans(s);
    set<tuple<int, int, int, int>> st;
    int ans = 0, nowx = 0, nowy = 0;
    for (const auto& aa : s) {
      if (aa == 'N') {
        if (st.find({nowx, nowx, nowy, nowy + 1}) != st.end())
          ans += 1;
        else
          ans += 5;
        st.insert({nowx, nowx, nowy, nowy + 1});
        ++nowy;
      } else if (aa == 'S') {
        if (st.find({nowx, nowx, nowy - 1, nowy}) != st.end())
          ans += 1;
        else
          ans += 5;
        st.insert({nowx, nowx, nowy - 1, nowy});
        --nowy;
      } else if (aa == 'E') {
        if (st.find({nowx, nowx + 1, nowy, nowy}) != st.end())
          ans += 1;
        else
          ans += 5;
        st.insert({nowx, nowx + 1, nowy, nowy});
        ++nowx;
      } else if (aa == 'W') {
        if (st.find({nowx - 1, nowx, nowy, nowy}) != st.end())
          ans += 1;
        else
          ans += 5;
        st.insert({nowx - 1, nowx, nowy, nowy});
        --nowx;
      }
    }
    printf("%d\n", ans);
  }
  return 0;
}
