The polar bears are going fishing. They plan to sail from (*s*<sub class="lower-index">*x*</sub>, *s*<sub class="lower-index">*y*</sub>) to (*e*<sub class="lower-index">*x*</sub>, *e*<sub class="lower-index">*y*</sub>). However, the boat can only sail by wind. At each second, the wind blows in one of these directions: east, south, west or north. Assume the boat is currently at (*x*, *y*).
 -  If the wind blows to the east, the boat will move to (*x* + 1, *y*). -  If the wind blows to the south, the boat will move to (*x*, *y* - 1). -  If the wind blows to the west, the boat will move to (*x* - 1, *y*). -  If the wind blows to the north, the boat will move to (*x*, *y* + 1). 
Alternatively, they can hold the boat by the anchor. In this case, the boat stays at (*x*, *y*). Given the wind direction for *t* seconds, what is the earliest time they sail to (*e*<sub class="lower-index">*x*</sub>, *e*<sub class="lower-index">*y*</sub>)?
The first line contains five integers *t*, *s*<sub class="lower-index">*x*</sub>, *s*<sub class="lower-index">*y*</sub>, *e*<sub class="lower-index">*x*</sub>, *e*<sub class="lower-index">*y*</sub> (1 ≤ *t* ≤ 10<sup class="upper-index">5</sup>,  - 10<sup class="upper-index">9</sup> ≤ *s*<sub class="lower-index">*x*</sub>, *s*<sub class="lower-index">*y*</sub>, *e*<sub class="lower-index">*x*</sub>, *e*<sub class="lower-index">*y*</sub> ≤ 10<sup class="upper-index">9</sup>). The starting location and the ending location will be different.

The second line contains *t* characters, the *i*-th character is the wind blowing direction at the *i*-th second. It will be one of the four possibilities: "E" (east), "S" (south), "W" (west) and "N" (north).
t,x1,y1,x2,y2=list(map(int,input().split()))
n=input()
p=''
k=0
c=''
u=0
if x1 > x2:
    p=p+'W'*(x1-x2)
    a='W'
else:
    p = p + 'E' * (x2 - x1)
    a='E'
if y1 > y2:
    p=p+'S'*(y1-y2)
    b='S'
else:
    p = p + 'N' * (y2 - y1)
    b='N'
for i in n:
    u+=1
    if i == a or i == b:
        c+=i
        if c.count(a) == p.count(a) and c.count(b) == p.count(a):
            print(u)
            k=1
            break
if k == 0:
    print(-1)


