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*.
import math,sys
from sys import stdin, stdout
from collections import Counter, defaultdict, deque
input = stdin.readline
I = lambda:int(input())
li = lambda:list(map(int,input().split()))


def case():
    a=input().strip()
    b=input().strip()
    d=deque(a)
    ans=[]
    i=0
    c=a.count('1')
    f=1
    while(True):
        if(b[i]=='1'):
            if(c%2):
                ans.append(1)
            else:
                while(len(d) and d[0]=='0'):
                    d.popleft()
                if(len(d)):
                    d.popleft()
                    ans.append('1')
                else:
                    f=0
                    break
        else:
            if(not c%2):
                ans.append(0)
            else:
                while(len(d) and d[0]=='0'):
                    d.popleft()
                if(len(d)):
                    d.popleft()
                    ans.append('0')
                    c-=1
                else:
                    f=0
                    break 
        #print(d)
        i+=1
        if(i==len(b)):
            break 
    
    ans="".join(map(str,ans))
    #print(ans)
    if(ans==b):
        print("YES")
    else:
        print("NO")
    



    

for _ in range(1):
    case()