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 N = 1e6 + 7;
const int M = 4e6 + 7;
const int lim = 1e6;
const int mod = 1e9 + 7;
const int inf = 0x3f3f3f3f;
const double eps = 1e-8;
const long long xxf = 1e18;
multiset<long long> ms;
priority_queue<int> pq;
void blue() { std::ios::sync_with_stdio(false), cin.tie(0), cout.tie(0); }
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 1) + (x << 3) + (ch ^ 48);
    ch = getchar();
  }
  return x * f;
}
int T;
char s[N];
map<pair<double, double>, int> mp;
int main() {
  blue();
  scanf("%d", &T);
  while (T--) {
    mp.clear();
    double x = 0, y = 0;
    scanf("%s", s);
    int ans = 0;
    pair<double, double> t;
    t.first = 0;
    t.second = 0;
    mp[t] = 1;
    int n = strlen(s);
    for (register int i = 0; i < n; i++) {
      pair<double, double> tt;
      if (s[i] == 'S') tt = pair<double, double>(x, y - 0.5), y -= 1.0;
      if (s[i] == 'N') tt = pair<double, double>(x, y + 0.5), y += 1.0;
      if (s[i] == 'W') tt = pair<double, double>(x - 0.5, y), x -= 1.0;
      if (s[i] == 'E') tt = pair<double, double>(x + 0.5, y), x += 1.0;
      if (mp[tt] == 1)
        ans += 1;
      else
        ans += 5, mp[tt] = 1;
    }
    printf("%d\n", ans);
  }
  return 0;
}
