[THE SxPLAY & KIVΛ - 漂流](https://soundcloud.com/kivawu/hyouryu)

[KIVΛ & Nikki Simmons - Perspectives](https://soundcloud.com/kivawu/perspectives)

With a new body, our idol Aroma White (or should we call her Kaori Minamiya?) begins to uncover her lost past through the OS space.

The space can be considered a 2D plane, with an infinite number of data nodes, indexed from 0, with their coordinates defined as follows:

  * The coordinates of the 0-th node is (x_0, y_0) 
  * For i > 0, the coordinates of i-th node is (a_x ⋅ x_{i-1} + b_x, a_y ⋅ y_{i-1} + b_y) 



Initially Aroma stands at the point (x_s, y_s). She can stay in OS space for at most t seconds, because after this time she has to warp back to the real world. She doesn't need to return to the entry point (x_s, y_s) to warp home.

While within the OS space, Aroma can do the following actions:

  * From the point (x, y), Aroma can move to one of the following points: (x-1, y), (x+1, y), (x, y-1) or (x, y+1). This action requires 1 second. 
  * If there is a data node at where Aroma is staying, she can collect it. We can assume this action costs 0 seconds. Of course, each data node can be collected at most once. 



Aroma wants to collect as many data as possible before warping back. Can you help her in calculating the maximum number of data nodes she could collect within t seconds?

Input

The first line contains integers x_0, y_0, a_x, a_y, b_x, b_y (1 ≤ x_0, y_0 ≤ 10^{16}, 2 ≤ a_x, a_y ≤ 100, 0 ≤ b_x, b_y ≤ 10^{16}), which define the coordinates of the data nodes.

The second line contains integers x_s, y_s, t (1 ≤ x_s, y_s, t ≤ 10^{16}) – the initial Aroma's coordinates and the amount of time available.

Output

Print a single integer — the maximum number of data nodes Aroma can collect within t seconds.

Examples

Input


1 1 2 3 1 0
2 4 20


Output


3

Input


1 1 2 3 1 0
15 27 26


Output


2

Input


1 1 2 3 1 0
2 2 1


Output


0

Note

In all three examples, the coordinates of the first 5 data nodes are (1, 1), (3, 3), (7, 9), (15, 27) and (31, 81) (remember that nodes are numbered from 0).

In the first example, the optimal route to collect 3 nodes is as follows: 

  * Go to the coordinates (3, 3) and collect the 1-st node. This takes |3 - 2| + |3 - 4| = 2 seconds. 
  * Go to the coordinates (1, 1) and collect the 0-th node. This takes |1 - 3| + |1 - 3| = 4 seconds. 
  * Go to the coordinates (7, 9) and collect the 2-nd node. This takes |7 - 1| + |9 - 1| = 14 seconds. 



In the second example, the optimal route to collect 2 nodes is as follows: 

  * Collect the 3-rd node. This requires no seconds. 
  * Go to the coordinates (7, 9) and collect the 2-th node. This takes |15 - 7| + |27 - 9| = 26 seconds. 



In the third example, Aroma can't collect any nodes. She should have taken proper rest instead of rushing into the OS space like that.
import sys 
# sys.setrecursionlimit(10**6) 
from sys import stdin, stdout
import bisect            #c++ upperbound
import math
import heapq
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 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
"""*******************************************************"""
def main():
    x,y,ax,ay,bx,by=cin()
    s=[(x,y)]
    for i in range(100):
        if(math.log10(ax)+math.log10(x)>18 or math.log10(ay)+math.log10(y)>18):
            break
        xx=ax*x+bx
        yy=ay*y+by
        x=xx
        y=yy
        s.append((x,y))
    x,y,t=cin()
    n=len(s)
    p=0
    if (x,y) in s:
        p=1
        s.remove((x,y))
    # print(s)
    n=len(s)
    an=p
    for i in range(n):
        f=0
        ans=p
        xx=x
        yy=y
        for j in range(i,n):
            f+=abs(s[j][0]-xx)+abs(s[j][1]-yy)
            xx=s[j][0]
            yy=s[j][1]
            if(f>t):
                break
            ans+=1
        an=max(an,ans)
        # print(ans)
    s=s[::-1]
    for i in range(n):
        f=0
        ans=p
        xx=x
        yy=y
        for j in range(i,n):
            f+=abs(s[j][0]-xx)+abs(s[j][1]-yy)
            xx=s[j][0]
            yy=s[j][1]
            if(f>t):
                break
            ans+=1
        an=max(an,ans)
    print(an)
######## 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'0' [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'0' [0]:
        A.append(sign*numb)
    return A
 
if __name__== "__main__":
  main()