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
t = int(input())

for it in range(t):
    alphas = input()

    x,y = 0,0
    total_time = 0

    path_d = dict()
    for a in alphas:
        if a == "S":
            forward = path_d.get((x, y, x, y-1), 5)
            back = path_d.get((x, y-1, x, y), 5)
            total_time += min(forward, back)
            path_d[(x, y-1, x, y)] = 1
            path_d[(x, y, x, y-1)] = 1
            (x, y) = (x, y-1)

        if a == "N":
            forward = path_d.get((x, y, x, y+1), 5)
            back = path_d.get((x, y+1, x, y), 5)
            total_time += min(forward, back)
            path_d[(x, y+1, x, y)] = 1
            path_d[(x, y, x, y+1)] = 1
            (x, y) = (x, y+1)

        if a == "W":
            forward = path_d.get((x, y, x-1, y), 5)
            back = path_d.get((x, y, x-1, y), 5)
            total_time += min(forward, back)
            path_d[(x-1, y, x, y)] = 1
            path_d[(x, y, x-1, y)] = 1
            (x, y) = (x-1, y)

        if a == "E":
            forward = path_d.get((x, y, x+1, y), 5)
            back = path_d.get((x+1, y, x, y), 5)
            total_time += min(forward, back)
            path_d[(x+1, y, x, y)] = 1
            path_d[(x, y, x+1, y)] = 1
            (x, y) = (x+1, y)
    print(total_time)
