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;
int dx8[] = {0, 0, 1, 1, 1, -1, -1, -1};
int dy8[] = {1, -1, 1, -1, 0, 0, -1, 1};
int dx4[] = {0, 0, 1, -1};
int dy4[] = {1, -1, 0, 0};
const double pi = acos(-1.0);
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  a %= b;
  return gcd(b, a);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  int test;
  cin >> test;
  while (test--) {
    string s;
    cin >> s;
    long long c = 0;
    map<pair<pair<int, int>, pair<int, int> >, int> mp;
    long long x = 0, y = 0;
    pair<int, int> pre, now;
    pre = {0, 0};
    for (int i = 0; i < s.size(); i++) {
      now = pre;
      if (s[i] == 'N')
        now.first++;
      else if (s[i] == 'S')
        now.first--;
      else if (s[i] == 'E')
        now.second--;
      else if (s[i] == 'W')
        now.second++;
      if (mp.find({now, pre}) != mp.end() or mp.find({pre, now}) != mp.end())
        c++;
      else {
        c += 5;
        mp[{now, pre}] = 1;
        mp[{pre, now}] = 1;
      }
      pre = now;
    }
    cout << c << endl;
    mp.clear();
  }
  return 0;
}
