Two people are playing a game with a string s, consisting of lowercase latin letters. 

On a player's turn, he should choose two consecutive equal letters in the string and delete them. 

For example, if the string is equal to "xaax" than there is only one possible turn: delete "aa", so the string will become "xx". A player not able to make a turn loses.

Your task is to determine which player will win if both play optimally.

Input

The only line contains the string s, consisting of lowercase latin letters (1 ≤ |s| ≤ 100 000), where |s| means the length of a string s.

Output

If the first player wins, print "Yes". If the second player wins, print "No".

Examples

Input


abacaba


Output


No


Input


iiq


Output


Yes


Input


abba


Output


No

Note

In the first example the first player is unable to make a turn, so he loses.

In the second example first player turns the string into "q", then second player is unable to move, so he loses.
from sys import stdin,stdout,setrecursionlimit
from functools import lru_cache, cmp_to_key
from heapq import merge, heapify, heappop, heappush,nlargest
from math import *
from collections import defaultdict as dd, deque, Counter as C
from itertools import combinations as comb, permutations as perm , accumulate
from bisect import bisect_left as bl, bisect_right as br, bisect
from time import perf_counter
from fractions import Fraction
import copy
import time
setrecursionlimit(10**9)
starttime = time.time()
mod = int(pow(10, 9) + 7)
mod2 = 998244353
# from sys import stdin
# input = stdin.readline
#def data(): return sys.stdin.readline().strip()
def data(): return input()
def num():return int(input())
def L(): return list(sp())
def LF(): return list(spf())
def sl(): return list(ssp())
def sp(): return map(int, data().split())
def spf(): return map(int, input.readline().split())
def ssp(): return map(str, data().split())
def l1d(n, val=0): return [val for i in range(n)]
def l2d(n, m, val=0): return [l1d(n, val) for j in range(m)]
def pmat(A):
    for ele in A:
        print(*ele,end="\n")
def pmat2(A):
    for ele in A:
        for j in ele:
            print(j,end='')
        print()
def iseven(n):
    return n%2==0

def seive(r):
    prime=[1 for i in range(r+1)]
    prime[0]=0
    prime[1]=0
    for i in range(r+1):
        if(prime[i]):
            for j in range(2*i,r+1,i):
                prime[j]=0
    return prime
#solution
#ACPC 
#remeber cut ribbon problem
# set data structures faster than binary search sometimes
#bipartite match dfs
#think in problems with recursive manner.
def solve(s):
    stack=[];ans=0
    for i in s:
        try:
            if stack[-1]==i:
                stack.pop()
                ans+=1
            else:
                stack.append(i)
        except:
            stack.append(i)
            continue
    return ans%2
string=input()
print("Yes" if solve(string) else "No")


endtime = time.time()
#print(f"Runtime of the program is {endtime - starttime}")