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 main() {
  long long t;
  cin >> t;
  string str;
  while (t--) {
    cin >> str;
    map<pair<long long, long long>, long long> m;
    pair<long long, long long> st = {0, 0};
    long long sum = 0;
    for (int i = 0; i < str.length(); i++)
      if (str[i] == 'N') {
        if (m[{st.first + 1, st.second}])
          sum++;
        else {
          m[{st.first + 1, st.second}] = 1;
          sum += 5;
        }
        st = {st.first + 2, st.second};
      } else if (str[i] == 'E') {
        if (m[{st.first, st.second - 1}])
          sum++;
        else {
          m[{st.first, st.second - 1}] = 1;
          sum += 5;
        }
        st = {st.first, st.second - 2};
      } else if (str[i] == 'S') {
        if (m[{st.first - 1, st.second}])
          sum++;
        else {
          m[{st.first - 1, st.second}] = 1;
          sum += 5;
        }
        st = {st.first - 2, st.second};
      } else if (str[i] == 'W') {
        if (m[{st.first, st.second + 1}])
          sum++;
        else {
          m[{st.first, st.second + 1}] = 1;
          sum += 5;
        }
        st = {st.first, st.second + 2};
      }
    cout << sum << "\n";
  }
}
