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, <image>. 
  * Remove the first character of a. For example, <image>. 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.

Input

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.

Output

Print "YES" (without quotes) if it is possible to turn a into b, and "NO" (without quotes) otherwise.

Examples

Input

01011
0110


Output

YES


Input

0011
1110


Output

NO

Note

In the first sample, the steps are as follows: 01011 → 1011 → 011 → 0110
//package Div_180;
//Xudo xoxlasa Accepted
import java.util.Scanner;

public class C {
public static void main(String[] args) {
	 Scanner sc=new Scanner(System.in);
	 String s=sc.next();
	 String s1=sc.next();
	 int bir1=0;
	 for (int i = 0; i <s.length(); i++) {
		bir1+=(s.charAt(i)-48);
	}
	 int bir2=0;
	 for (int i = 0; i <s1.length(); i++) {
		 bir2+=(s1.charAt(i)-48);
	 }
	 if(bir1==bir2){
		 System.out.println("YES");
	 }else{
		 if(bir1<bir2){
			 if(bir1%2==0){
				 System.out.println("NO");
			 }else{
				 bir1++;
				 if(bir1==bir2){
					 System.out.println("YES");
				 }else{
					 System.out.println("NO");
				 }
			 }
		 }else{
			 System.out.println("YES");
		 }
	 }
}
}
