After having a wonderful time at the Carnival test festival, Jão and his friends decided to go for an after party in his house. At some point Jão convinced them to play a classical mathematical puzzle. He put his friends in a line and gave each one of them a typical Carnival hat colored in black or white. They were arranged in such a way that they could only see the hats of people that were in front of them. In particular, the last person could see everyone's hat and the first could see nothing.
This game must be played from the last one on the line to the first. Each person should try, in this order, to guess the color of his own hat out loud. For that, the participants can agree on some strategy before getting the hats.
Jão had so much fun watching his drunk friends trying to tackle the puzzle. Now he decided to teach them a powerful strategy. Let nn be the number of participants in the game, and let they have numbers from 1 to nn, from the first person in line to the last. Let the black hat have an associated value of 11 and the white one have a value of 00. Then we can always guess at least n−1n−1 values correctly by following this strategy:
- The last one in line have no information about his own hat, so he should guess the xor (⊕⊕) of all the hats he can see;
- The others players in line should do something different. Let XXbe the xor of the values guessed by previous players. Let YY be the xor of the hats the current player can see. If X=YX=Y, then his own hat must be white, otherwise it must be black;
Notice that every player will follow (2), except for the last one in line, which will follow (1). This strategy ensures that the first n−1n−1 participants in line will guess correctly. Can you see why it works?
Unfortunately, Jão's friends are drunk, so they tend to mess the strategy up. When someone messes up, that means that he should have guessed xx according to the strategy, but guessed x⊕1x⊕1. Your task, given the guesses of each participant in line, is to decide if there is a hat distribution such that we can say no one messed up given those guesses.
Input
The only input line contains a binary string ss (|s|=n≤105|s|=n≤105). The ii-th character of ss represents the color guessed by the ii-th guy in the line (0 for white, 1 for black).
Output
Output YES if there is a hat distribution such that we can say no one messed up. Otherwise, output NO.
Examples
Input
10010
Output
YES
Input
01110
Output
NO
#include <iostream>
#include<stdio.h>
#include<string.h>
#include<set>
using namespace std;
const int N=1e5+10;
int main()
{
char s[N];
scanf("%s",s);
int n=strlen(s);
int k=s[0]-'0';
for(int i=1; i<n-1; i++)
{
k^=s[i]-'0';
}
int p=s[n-1]-'0';
if(k==p)
printf("YES\n");
else printf("NO\n");
return 0;
}