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*.
alice = input()
bob = input()

a = alice.count('1') 
b = bob.count('1')
a += (a%2) # if it is odd, then add another one.

print("YES" if a >= b else "NO")

