You a captain of a ship. Initially you are standing in a point (x_1, y_1) (obviously, all positions in the sea can be described by cartesian plane) and you want to travel to a point (x_2, y_2). 

You know the weather forecast — the string s of length n, consisting only of letters U, D, L and R. The letter corresponds to a direction of wind. Moreover, the forecast is periodic, e.g. the first day wind blows to the side s_1, the second day — s_2, the n-th day — s_n and (n+1)-th day — s_1 again and so on. 

Ship coordinates change the following way:

  * if wind blows the direction U, then the ship moves from (x, y) to (x, y + 1); 
  * if wind blows the direction D, then the ship moves from (x, y) to (x, y - 1); 
  * if wind blows the direction L, then the ship moves from (x, y) to (x - 1, y); 
  * if wind blows the direction R, then the ship moves from (x, y) to (x + 1, y). 



The ship can also either go one of the four directions or stay in place each day. If it goes then it's exactly 1 unit of distance. Transpositions of the ship and the wind add up. If the ship stays in place, then only the direction of wind counts. For example, if wind blows the direction U and the ship moves the direction L, then from point (x, y) it will move to the point (x - 1, y + 1), and if it goes the direction U, then it will move to the point (x, y + 2).

You task is to determine the minimal number of days required for the ship to reach the point (x_2, y_2).

Input

The first line contains two integers x_1, y_1 (0 ≤ x_1, y_1 ≤ 10^9) — the initial coordinates of the ship.

The second line contains two integers x_2, y_2 (0 ≤ x_2, y_2 ≤ 10^9) — the coordinates of the destination point.

It is guaranteed that the initial coordinates and destination point coordinates are different.

The third line contains a single integer n (1 ≤ n ≤ 10^5) — the length of the string s.

The fourth line contains the string s itself, consisting only of letters U, D, L and R.

Output

The only line should contain the minimal number of days required for the ship to reach the point (x_2, y_2).

If it's impossible then print "-1".

Examples

Input


0 0
4 6
3
UUU


Output


5


Input


0 3
0 0
3
UDD


Output


3


Input


0 0
0 1
1
L


Output


-1

Note

In the first example the ship should perform the following sequence of moves: "RRRRU". Then its coordinates will change accordingly: (0, 0) → (1, 1) → (2, 2) → (3, 3) → (4, 4) → (4, 6).

In the second example the ship should perform the following sequence of moves: "DD" (the third day it should stay in place). Then its coordinates will change accordingly: (0, 3) → (0, 3) → (0, 1) → (0, 0).

In the third example the ship can never reach the point (0, 1).
from __future__ import division, print_function
# import threading
# threading.stack_size(2**27)
# import sys
# sys.setrecursionlimit(10**7)
from sys import stdin, stdout
import bisect            #c++ upperbound
import math
import heapq
i_m=9223372036854775807
def modinv(n,p):
    return pow(n,p-2,p)
def cin():
    return map(int,sin().split())
def ain():                           #takes array as input
    return list(map(int,sin().split()))
def sin():
    return input()
def inin():
    return int(input())
import math  
def GCD(x, y): 
    x=abs(x)
    y=abs(y)
    if(min(x,y)==0):
        return max(x,y)
    while(y): 
        x, y = y, x % y 
    return x 
def Divisors(n) : 
    l = []  
    for i in range(1, int(math.sqrt(n) + 1)) :
        if (n % i == 0) : 
            if (n // i == i) : 
                l.append(i) 
            else : 
                l.append(i)
                l.append(n//i)
    return l
prime=[]
def SieveOfEratosthenes(n): 
    global prime
    prime = [True for i in range(n+1)] 
    p = 2
    while (p * p <= n): 
        if (prime[p] == True): 
            for i in range(p * p, n+1, p): 
                prime[i] = False
        p += 1
    f=[]
    for p in range(2, n): 
        if prime[p]: 
            f.append(p)
    return f
q=[]       
def dfs(n,d,v,c):
    global q
    v[n]=1
    x=d[n]
    q.append(n)
    j=c
    for i in x:
        if i not in v:
            f=dfs(i,d,v,c+1)
            j=max(j,f)
            # print(f)
    return j

def fun(a,b,c,d):
    p=(c-a+1)*(d-b+1)//2;
    if((a%2)==(b%2) and ((c-a+1)&1) and ((d-b+1)&1)):
        p+=1
    return p
def ch(m,p,x,y):
    t=len(p)
    # print(m)
    mm=m
    f=m%t
    m=m//t
    j=p[-1][0]*m
    k=p[-1][1]*m
    x-=j
    y-=k
    if(f>0):
        f-=1
        j=p[f][0]
        k=p[f][1]
        x-=j
        y-=k
    # print(p,f,m,x,y,abs(x)+abs(y),mm)
    if(abs(x)+abs(y)<=mm):
        return 1
"""*******************************************************"""
def main():
    x,y=cin()
    a,b=cin()
    x=a-x
    y=b-y
    a=b=0
    n=inin()
    s=sin()
    p=[]
    for i in s:
        if i=="U":
            b+=1
        if i=="D":
            b-=1
        if i=="L":
            a-=1
        if i=="R":
            a+=1
        p.append((a,b))
    l=1
    r=i_m
    ans=-1
    while(r>l):
        m=l+(r-l)//2
        if(ch(m,p,x,y)):
            ans=m
            # print(ans)
            r=m
        else:
            l=m+1
    print(ans)

    
######## Python 2 and 3 footer by Pajenegod and c1729
 
# Note because cf runs old PyPy3 version which doesn't have the sped up
# unicode strings, PyPy3 strings will many times be slower than pypy2.
# There is a way to get around this by using binary strings in PyPy3
# but its syntax is different which makes it kind of a mess to use.
 
# So on cf, use PyPy2 for best string performance.
 
py2 = round(0.5)
if py2:
    from future_builtins import ascii, filter, hex, map, oct, zip
    range = xrange
 
import os, sys
from io import IOBase, BytesIO
 
BUFSIZE = 8192
class FastIO(BytesIO):
    newlines = 0
 
    def __init__(self, file):
        self._file = file
        self._fd = file.fileno()
        self.writable = "x" in file.mode or "w" in file.mode
        self.write = super(FastIO, self).write if self.writable else None
 
    def _fill(self):
        s = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
        self.seek((self.tell(), self.seek(0,2), super(FastIO, self).write(s))[0])
        return s
 
    def read(self):
        while self._fill(): pass
        return super(FastIO,self).read()
 
    def readline(self):
        while self.newlines == 0:
            s = self._fill(); self.newlines = s.count(b"\n") + (not s)
        self.newlines -= 1
        return super(FastIO, self).readline()
 
    def flush(self):
        if self.writable:
            os.write(self._fd, self.getvalue())
            self.truncate(0), self.seek(0)
 
class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        if py2:
            self.write = self.buffer.write
            self.read = self.buffer.read
            self.readline = self.buffer.readline
        else:
            self.write = lambda s:self.buffer.write(s.encode('ascii'))
            self.read = lambda:self.buffer.read().decode('ascii')
            self.readline = lambda:self.buffer.readline().decode('ascii')
 
 
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip('\r\n')
 
# Cout implemented in Python
import sys
class ostream:
    def __lshift__(self,a):
        sys.stdout.write(str(a))
        return self
cout = ostream()
endl = '\n'
 
# Read all remaining integers in stdin, type is given by optional argument, this is fast
def readnumbers(zero = 0):
    conv = ord if py2 else lambda x:x
    A = []; numb = zero; sign = 1; i = 0; s = sys.stdin.buffer.read()
    try:
        while True:
            if s[i] >= b'R' [0]:
                numb = 10 * numb + conv(s[i]) - 48
            elif s[i] == b'-' [0]: sign = -1
            elif s[i] != b'\r' [0]:
                A.append(sign*numb)
                numb = zero; sign = 1
            i += 1
    except:pass
    if s and s[-1] >= b'R' [0]:
        A.append(sign*numb)
    return A
 
# threading.Thread(target=main).start()
if __name__== "__main__":
  main()