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*.
def maxConsecutive(a):
    i = 1
    curr = 0
    if a[0] == '1':
        curr += 1
    ans = curr
    for i in range(len(a)):
        if a[i] == '1':
            curr += 1
        else:
            curr = 1
        ans = max(ans, curr)
    return ans

def countOne(a):
    ans = 0
    for i in range(len(a)):
        if a[i] == '1':
            ans += 1
    return ans

a = input()
b = input()
if countOne(a) + countOne(a)%2 >= countOne(b):
    print("YES")
else:
    print("NO")

# 101101
#       01111