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;
long long MOD = 1000000007;
long long INF = 1000000000000000000;
long long powMod(long long x, long long n, long long mod) {
  if (n == 0) return 1;
  long long t = powMod(x, n / 2, mod);
  t = t * t % mod;
  if (n & 1) return t * x % mod;
  return t;
}
long long gcd(long long a, long long b) {
  if (a == 0 || b == 0) return a + b;
  if (b > a) return gcd(b, a);
  return gcd(b, a % b);
}
int main(void) {
  int t;
  string s;
  scanf("%d", &t);
  set<tuple<int, int, int, int>> e;
  for (int i = 0; i < int(t); i++) {
    e.clear();
    cin >> s;
    int x = 0, y = 0, dx, dy;
    int ans = 0;
    for (int j = 0; j < int(s.size()); j++) {
      if (s[j] == 'N') dx = 0, dy = 1;
      if (s[j] == 'S') dx = 0, dy = -1;
      if (s[j] == 'E') dx = 1, dy = 0;
      if (s[j] == 'W') dx = -1, dy = 0;
      if (e.find(make_tuple(x, y, x + dx, y + dy)) != e.end())
        ans += 1;
      else {
        ans += 5;
        e.insert(make_tuple(x, y, x + dx, y + dy));
        e.insert(make_tuple(x + dx, y + dy, x, y));
      }
      x += dx;
      y += dy;
    }
    printf("%d\n", ans);
  }
  return 0;
}
