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;
struct node {
  int x, y;
  int dir;
};
bool operator<(node x, node y) {
  if (x.x != y.x) return x.x < y.x;
  if (x.y != y.y) return x.y < y.y;
  if (x.dir != y.dir) return x.dir < y.dir;
  return false;
}
int n, len, x, y, ans;
char s[200005];
map<node, bool> mp;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    mp.clear();
    ans = x = y = 0;
    scanf("%s", s + 1);
    len = strlen(s + 1);
    for (int j = 1; j <= len; j++) {
      ans += 5;
      if (s[j] == 'N') {
        if (mp[(node){x, y, 0}]) ans -= 4;
        mp[(node){x, y, 0}] = true;
        mp[(node){x + 1, y, 1}] = true;
        x++;
      }
      if (s[j] == 'S') {
        if (mp[(node){x, y, 1}]) ans -= 4;
        mp[(node){x, y, 1}] = true;
        mp[(node){x - 1, y, 0}] = true;
        x--;
      }
      if (s[j] == 'W') {
        if (mp[(node){x, y, 3}]) ans -= 4;
        mp[(node){x, y, 3}] = true;
        mp[(node){x, y + 1, 4}] = true;
        y++;
      }
      if (s[j] == 'E') {
        if (mp[(node){x, y, 4}]) ans -= 4;
        mp[(node){x, y, 4}] = true;
        mp[(node){x, y - 1, 3}] = true;
        y--;
      }
    }
    printf("%d\n", ans);
  }
}
