You are fishing with polar bears Alice and Bob. While waiting for the fish to bite, the polar bears get bored. They come up with a game. First Alice and Bob each writes a 01-string (strings that only contain character "0" and "1") *a* and *b*. Then you try to turn *a* into *b* using two types of operations:
 -  Write *parity*(*a*) to the end of *a*. For example, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/2489b4bd10b8f67bed8ee4614ff21d963da20895.png" style="max-width: 100.0%;max-height: 100.0%;"/>. -  Remove the first character of *a*. For example, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/073a30771e84caf65e9a79493bf290f14d315cbe.png" style="max-width: 100.0%;max-height: 100.0%;"/>. You cannot perform this operation if *a* is empty. 
You can use as many operations as you want. The problem is, is it possible to turn *a* into *b*?

The *parity* of a 01-string is 1 if there is an odd number of "1"s in the string, and 0 otherwise.
The first line contains the string *a* and the second line contains the string *b* (1 ≤ |*a*|, |*b*| ≤ 1000). Both strings contain only the characters "0" and "1". Here |*x*| denotes the length of the string *x*.
from bisect import *        
from sys import *
from time import *
start = time()

T = input()
S = []

for i, t in  enumerate(T):
    if ord(t) >= 97 and ord(t) <= 122:
        S += [t]

L = [[] for i in range(len(S))]
R = [[] for i in range(len(S))]

for i in range(len(S)):
    ind = 0
    while i + ind < len(S) and i - ind >= 0:
        if S[i + ind] == S[i - ind]:
            L[i - ind].append(1 + 2 * ind)
            R[i + ind].append(1 + 2 * ind)
        else:
            break
        ind += 1

for i in range(1, len(S)):
    ind = 0
    while i - 1 - ind >= 0 and i + ind < len(S):
        if S[i - 1 - ind] == S[i + ind]:
            L[i - 1 - ind].append(2 + 2 * ind)
            R[i + ind].append(2 + 2 * ind)
        else:
            break
        ind += 1
print(time() - start)    
for l in L:
    l.sort()
for r in R:
    r.sort()
print(time() - start)

print(L)
print(R)

ans = 0
ind = 0

while ind < len(S):
    l = ind
    r = l + 1
    while r < len(S) and S[r] == S[l]:
        r += 1
    f



print(time() - start)     
        
print(ans) 

