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*.
a=input()
b=input()
k=0
for i in range(len(b)-1,-1,-1):
    if(b[i]=='1'):
        k=i
        break
c=b[0:k+1]
r= [i for i in range(len(a)) if a.startswith(c, i)]
if(c not in a):
    print("NO")
    quit()
j=max(r)
d=a[j+len(c):]
if('1' in d):
    print("NO")
else:
    z=b[k+1:]
    if(len(d)<=len(z)):
        print("YES")
    else:
        print("NO")

