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 int MOD = 1000000007;
long long int gcd(long long int a, long long int b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
long long int lcm(long long int a, long long int b) {
  return a / gcd(a, b) * b;
}
long long int power(long long int a, long long int n) {
  if (n == 0) return 1;
  long long int p = power(a, n / 2);
  p = (p * p);
  if (n % 2)
    return (p * a);
  else
    return p;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int test = 1;
  cin >> test;
  while (test--) {
    string s;
    cin >> s;
    map<pair<pair<int, int>, pair<int, int>>, int> m;
    pair<int, int> prev = {0, 0};
    pair<int, int> curr = {0, 0};
    for (auto x : s) {
      if (x == 'N') {
        curr.second++;
      } else if (x == 'S') {
        curr.second--;
      } else if (x == 'W') {
        curr.first--;
      } else {
        curr.first++;
      }
      if (prev > curr)
        m[{curr, prev}]++;
      else
        m[{prev, curr}]++;
      prev = curr;
    }
    long long int ans = 0;
    for (auto x : m) {
      ans += (x.second - 1) + 5;
    }
    cout << ans << '\n';
  }
  return 0;
}
