题干：

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:

  • Writeparity(a) to the end ofa. For example,.
  • Remove the first character ofa. For example,. You cannot perform this operation ifais empty.

You can use as many operations as you want. The problem is, is it possible to turn ainto 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

题目大意：

  告诉你两种操作，问你能否将A串变成B串。操作1：删除第一个字符。操作2：在字符串后面添parity(a)，这个函数的值为0或1，分别在 串中有偶数个1，奇数个1  时取到。

解题报告：

  猜结论的思维题。。。模拟了一下发现真的可以，，。也就是看字符串中最多能造出多少个1来。，因为如果1 的数量大于等于b串中1的数量，，就一定能构造出来、、就是忘了如果1的个数为奇数的时候，，我们可以直接在后面添加一个1变成多一个1，

AC代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define ll long long
#define pb push_back
#define pm make_pair
#define fi first
#define se second
using namespace std;
const int MAX = 2e6 + 5;
int n,k;
char s[MAX],t[MAX];
int main()
{
  scanf("%s",s+1);
  scanf("%s",t+1);
  int cnt1=0,cnt2=0;
  for(int i = 1; i<=strlen(s+1); i++) {
    if(s[i] == '1') cnt1++;
  }
  for(int i = 1; i<=strlen(t+1); i++) {
    if(t[i] == '1') cnt2++;
  }
  if(cnt1&1) cnt1++;
  if(cnt1 >= cnt2) puts("YES");
  else puts("NO");
  return 0 ;
 }