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 long double PI = atan(1.0) * 4;
const int64_t INF = 8e18;
const int32_t LOG = 21;
const long long int MOD = 998244353;
int32_t main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  long long int tests = 1;
  cin >> tests;
  for (long long int ii = 0; ii < tests; ii++) {
    map<pair<pair<long long int, long long int>,
             pair<long long int, long long int> >,
        bool>
        m;
    string second;
    cin >> second;
    long long int cost = 0;
    long long int x = 0, y = 0;
    for (auto it : second) {
      long long int x1 = x, y1 = y;
      if (it == 'N')
        y1++;
      else if (it == 'S')
        y1--;
      else if (it == 'E')
        x1++;
      else
        x1--;
      if (m[{{x, y}, {x1, y1}}])
        cost += 1;
      else
        cost += 5;
      m[{{x, y}, {x1, y1}}] = m[{{x1, y1}, {x, y}}] = true;
      x = x1, y = y1;
    }
    cout << cost << '\n';
  }
  return 0;
}
