Solve in Python:
One of Arkady's friends works at a huge radio telescope. A few decades ago the telescope has sent a signal $s$ towards a faraway galaxy. Recently they've received a response $t$ which they believe to be a response from aliens! The scientists now want to check if the signal $t$ is similar to $s$.

The original signal $s$ was a sequence of zeros and ones (everyone knows that binary code is the universe-wide language). The returned signal $t$, however, does not look as easy as $s$, but the scientists don't give up! They represented $t$ as a sequence of English letters and say that $t$ is similar to $s$ if you can replace all zeros in $s$ with some string $r_0$ and all ones in $s$ with some other string $r_1$ and obtain $t$. The strings $r_0$ and $r_1$ must be different and non-empty.

Please help Arkady's friend and find the number of possible replacements for zeros and ones (the number of pairs of strings $r_0$ and $r_1$) that transform $s$ to $t$.


-----Input-----

The first line contains a string $s$ ($2 \le |s| \le 10^5$) consisting of zeros and ones — the original signal.

The second line contains a string $t$ ($1 \le |t| \le 10^6$) consisting of lowercase English letters only — the received signal.

It is guaranteed, that the string $s$ contains at least one '0' and at least one '1'.


-----Output-----

Print a single integer — the number of pairs of strings $r_0$ and $r_1$ that transform $s$ to $t$.

In case there are no such pairs, print $0$.


-----Examples-----
Input
01
aaaaaa

Output
4

Input
001
kokokokotlin

Output
2



-----Note-----

In the first example, the possible pairs $(r_0, r_1)$ are as follows:  "a", "aaaaa"  "aa", "aaaa"  "aaaa", "aa"  "aaaaa", "a"  

The pair "aaa", "aaa" is not allowed, since $r_0$ and $r_1$ must be different.

In the second example, the following pairs are possible:   "ko", "kokotlin"  "koko", "tlin"
import sys
from math import *

def minp():
	return sys.stdin.readline().strip()

def mint():
	return int(minp())

def mints():
	return map(int, minp().split())

def add(a,b):
	return (a+b)%1000000007

def sub(a,b):
	return (a+1000000007-b)%1000000007

def mul(a,b):
	return (a*b)%1000000007

p = 102367
s = list(map(int,minp()))
t = list(map(ord,minp()))
h = [0]*(len(t)+1)
pp = [1]*(len(t)+1)
for i in range(len(t)):
	h[i+1] = add(mul(h[i], p), t[i])
	pp[i+1] = mul(pp[i], p)

def cmp(a, b, l):
	if a > b:
		a, b = b, a
	h1 = sub(h[a+l], mul(h[a], pp[l]))
	h2 = sub(h[b+l], mul(h[b], pp[l]))
	return h2 == h1

c = [0,0]
idx = [-1,-1]
for i in range(len(s)):
	c[s[i]] += 1
	if idx[s[i]] < 0:
		idx[s[i]] = i
Mv = max(c)
mv = min(c)
Mi = c.index(Mv)
mi = (Mi^1)
lt = len(t)
sp = [0,0]
res = 0
for k in range(1,lt//Mv+1):
	l = [0,0]
	x = (lt-k*Mv)//mv
	if x > 0 and x*mv + k*Mv == lt:
		l[Mi] = k
		l[mi] = x
		if idx[0] < idx[1]:
			sp[0] = 0
			sp[1] = idx[1]*l[0]
		else:
			sp[1] = 0
			sp[0] = idx[0]*l[1]
		ok = True
		j = 0
		for i in range(len(s)):
			if not cmp(sp[s[i]], j, l[s[i]]):
				ok = False
				break
			j += l[s[i]]
		if l[0] == l[1] and cmp(sp[0], sp[1], l[0]):
			ok = False
		if ok:
			res += 1
print(res)