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
from math import *
from collections import *
from operator import itemgetter
import bisect
i = lambda: input()
ii = lambda: int(input())
iia = lambda: list(map(int,input().split()))
isa = lambda: list(input().split())
I = lambda:list(map(int,input().split()))
chrIdx = lambda x: ord(x)-96
idxChr = lambda x: chr(96+x)

t = ii()
for _ in range(t):
    s = input()
    d = defaultdict(lambda: [0,0,0,0])
    dm = {'N':0,'S':1,'E':2,'W':3}
    x,y = 0,0
    cost = 0
    for i in s:
        p = (x,y)
        if(i=='N'):
            y+=1
        elif(i=='S'):
            y-=1
        elif(i=='E'):
            x+=1
        elif(i=='W'):
            x-=1
        b = (x,y)
        ans1 = d[b]
        ans2 = d[p]
        if(i=='N'):
            if(d[p][0]==0):
                cost+=5
            else:
                cost+=1
            d[p][0] = 1
            d[b][1] = 1
        elif(i=='S'):
            if(d[p][1]==0):
                cost+=5
            else:
                cost+=1
            d[p][1] = 1
            d[b][0] = 1
        elif(i=='E'):
            if(d[p][2]==0):
                cost+=5
            else:
                cost+=1
            d[p][2] = 1
            d[b][3] = 1
        elif(i=='W'):
            if(d[p][3]==0):
                cost+=5
            else:
                cost+=1
            d[p][3] = 1
            d[b][2] = 1
    print(cost)
