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(int argc, char** argv) {
  int t, x, y;
  string line;
  map<pair<pair<int, int>, char>, bool> m;
  cin >> t;
  while (t--) {
    cin >> line;
    int l = line.length(), sum;
    sum = x = y = 0;
    m.clear();
    for (int i = 0; i < l; i++) {
      int temX = x, temY = y;
      if (line[i] == 'N') {
        temY++;
        if (m.find({{x, y}, 'N'}) == m.end()) {
          m[{{x, y}, 'N'}] = m[{{temX, temY}, 'S'}] = 1;
          sum += 5;
        } else
          sum += 1;
      } else if (line[i] == 'S') {
        temY--;
        if (m.find({{x, y}, 'S'}) == m.end()) {
          m[{{x, y}, 'S'}] = m[{{temX, temY}, 'N'}] = 1;
          sum += 5;
        } else
          sum += 1;
      } else if (line[i] == 'E') {
        temX++;
        if (m.find({{x, y}, 'E'}) == m.end()) {
          m[{{x, y}, 'E'}] = m[{{temX, temY}, 'W'}] = 1;
          sum += 5;
        } else
          sum += 1;
      } else {
        temX--;
        if (m.find({{x, y}, 'W'}) == m.end()) {
          m[{{x, y}, 'W'}] = m[{{temX, temY}, 'E'}] = 1;
          sum += 5;
        } else
          sum += 1;
      }
      x = temX, y = temY;
    }
    cout << sum << '\n';
  }
  return 0;
}
