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, .
• Remove the first character of a. For example, . 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 ainto b?

The parity

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

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

題解：按這兩種操作，是否可以把a串轉化為b串

Write parity(a) to the end of a. For example, .

Remove the first character of a. For example, . You cannot perform this operation if a is empty.題解：

我們可以發現當字串起始1個個數為奇數時，我們只能新增1，為偶數時可以新增任何數量的0，我們又可以刪去左邊的數，當刪去一個1的時候，右邊又可以新增1，這個過程可以說明，串a的1若為偶數num，可以轉化為1數量相同的任意串，若為奇數num，可以轉化1的數量為num或num+1的任何串，所以統計兩個串的1的數量比較即可。

#include<bits/stdc++.h>
using namespace std;
const int N=2*1e6+10;
int n,m,k;
char a[N],b[N];
int main()
{
	while(~scanf("%s%s",a+1,b+1))
	{
		int l1=strlen(a+1);
		int l2=strlen(b+1);
		int s1=0,s2=0;
		for(int i=1;i<=l1;i++)
		{
			if(a[i]=='1') s1++;
		}
		for(int j=1;j<=l2;j++)
		{
			if(b[j]=='1') s2++;
		}
		if(s1&1) s1++;
	//	cout<<s1<<" "<<s2<<endl;
		if(s1>=s2) cout<<"YES"<<endl;
		else cout<<"NO"<<endl;
	}
	return 0;
}