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
FAST_IO = 0
if FAST_IO:
    import io, sys, atexit
    rr = iter(sys.stdin.read().splitlines()).next
    sys.stdout = _OUTPUT_BUFFER = io.BytesIO()
    @atexit.register
    def write():
        sys.__stdout__.write(_OUTPUT_BUFFER.getvalue())
else:
    rr = raw_input
rri = lambda: int(rr())
rrm = lambda: map(int, rr().split())

####

def solve(S):
    def canonical(x1,y1,x2,y2):
        if (x1,y1) > (x2,y2):
            x1,y1,x2,y2=x2,y2,x1,y1
        return x1,y1,x2,y2
    seen = set() #set([(0,0)])
    ans = 0
    x=y=0
    for c in S:
        ox,oy = x,y
        if c == 'N':
            y += 1
        elif c == 'S':
            y -= 1
        elif c == 'W':
            x -= 1
        else:
            x += 1
        cc = canonical(ox,oy,x,y)
        if cc in seen:
            ans += 1
        else:
            ans += 5
        seen.add(cc)
    return ans

for _ in xrange(rri()):
    print solve(rr())
