【CodeForces - 298C】Parity Game （思维，有坑）

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:

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 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 ;}

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【CodeForces - 298C】Parity Game （思维，有坑）

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