1056_E. Check Transcription
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 ≤ |s| ≤ 10^5) consisting of zeros and ones — the original signal.

The second line contains a string t (1 ≤ |t| ≤ 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" 
```python
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)
```