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
# May the Speedforce be with us...
'''
for _ in range(int(input())):
arr=list(map(int,input().split()))
n,k=map(int,input().split())
n=int(input())
s=input()

from collections import defaultdict
d=defaultdict()

d=dict()
s=set()
s.intersection()
s.union()
s.difference()


problem statement achhe se padhna hai
age se bhi dekhna hai, piche se bhi dekhna hai
'''
from math import gcd,ceil
from collections import defaultdict as dd
def lcm(a,b):
	return (a*b)//gcd(a,b)
def inin():
	return int(input())
def inar():
	return list(map(int,input().split()))
def ar(element,size):
	return [element for i in range(size)]
def digitsum(num):
	su=0
	while(num):
		su+=num%10
		num//=10
	return su
def pred(n,k,pointer):
	return (pointer-k)>=n



for _ in range(inin()):
	vis=dd(bool)
	x1,y1=0,0
	s=input()
	su=0

	
	for i in s:
		x,y=x1,y1
		if i=='N':
			y+=1

		if i=='S':
			y-=1

		if i=='W':
			x-=1


		if i=='E':
			x+=1

		if vis[(x1,y1,x,y)]:
			su+=1
		else:
			vis[(x,y,x1,y1)]=True
			vis[(x1,y1,x,y)]=True
			su+=5
		x1=x
		y1=y
	print(su)