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 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

唉，那個奇偶的地方真的讓人煩死了。直接上程式碼

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;

const int maxx=1e3+10;
char s1[maxx];
char s2[maxx];

int main()
{
	while(cin>>s1>>s2)
	{
		int len1=strlen(s1);
		int len2=strlen(s2);
		int sum1=0;
		int sum2=0;
		for(int i=0;i<len1;i++) if(s1[i]=='1') sum1++;
		for(int i=0;i<len2;i++) if(s2[i]=='1') sum2++;
		if(sum1%2) sum1+=1;
		if(sum1>=sum2) cout<<"YES"<<endl;
		else cout<<"NO"<<endl;
	}
 } 

努力加油a啊，(^o)/~