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;
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization("O3")
#pragma GCC optimization("unroll-loops")
int del_x[4] = {-1, 1, 0, 0};
int del_y[4] = {0, 0, 1, -1};
int XX[8] = {-1, -1, -1, 0, 1, 1, 1, 0};
int YY[8] = {-1, 0, 1, 1, 1, 0, -1, -1};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    string s;
    cin >> s;
    long long int cost = 0;
    map<pair<pair<int, int>, pair<int, int>>, bool> m;
    int x = 0, y = 0;
    pair<int, int> p1;
    p1 = {0, 0};
    pair<int, int> p2;
    for (auto ch : s) {
      if (ch == 'N') {
        p2 = {++x, y};
      } else if (ch == 'S') {
        p2 = {--x, y};
      } else if (ch == 'W') {
        p2 = {x, --y};
      } else {
        p2 = {x, ++y};
      }
      if (m.count({p1, p2}) == 0) {
        cost += 5;
        m[{p1, p2}] = true;
        m[{p2, p1}] = true;
      } else {
        cost += 1;
      }
      p1 = p2;
    }
    cout << cost << "\n";
  }
}
