Problem:
Score : 700 points

Problem Statement

We have a string s consisting of lowercase English letters.
Snuke is partitioning s into some number of non-empty substrings.
Let the subtrings obtained be s_1, s_2, ..., s_N from left to right. (Here, s = s_1 + s_2 + ... + s_N holds.)
Snuke wants to satisfy the following condition:

For each i (1 \leq i \leq N), it is possible to permute the characters in s_i and obtain a palindrome.

Find the minimum possible value of N when the partition satisfies the condition.

Constraints

1 \leq |s| \leq 2 \times 10^5

s consists of lowercase English letters.

Input

Input is given from Standard Input in the following format:

s

Output

Print the minimum possible value of N when the partition satisfies the condition.

Sample Input 1

aabxyyzz

Sample Output 1

2

The solution is to partition s as aabxyyzz = aab + xyyzz.
Here, aab can be permuted to form a palindrome aba, and xyyzz can be permuted to form a palindrome zyxyz.

Sample Input 2

byebye

Sample Output 2

1

byebye can be permuted to form a palindrome byeeyb.

Sample Input 3

abcdefghijklmnopqrstuvwxyz

Sample Output 3

26

Sample Input 4

abcabcxabcx

Sample Output 4

3

The solution is to partition s as abcabcxabcx = a + b + cabcxabcx.

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#include <stdio.h>
using namespace std;

#define INF 2147483647
#define LINF 9223372036854775807
#define FOR(i,a,b) for (int i = (a); i < (b); ++i)
#define REP(i,n) FOR(i,0,n)
#define ALL(v) (v).begin(), (v).end()

typedef long long ll;
typedef pair<ll, int> P;

string s;
int cnt[26][200000],cnt2[26];
int ans;
bool flg[26],tmp;
int fst,snd;

int main(){
	cin.tie(0);
	ios::sync_with_stdio(false);

	cin>>s;
	REP(i,s.size()) {
		if(i!=0){
			REP(j,26) cnt[j][i]=cnt[j][i-1];
		}
		cnt[s[i]-'a'][i]++;
		cnt2[s[i]-'a']++;
	}


	bool ok=false;
	REP(i,26) {
		if(cnt2[i]%2==1) ok=true;
	}
	if(!ok) {
		cout<<'1'<<endl;
		return 0;
	}

	REP(i,26) flg[i]=false;

	REP(i,s.size()) {
		if(cnt2[s[i]-'a']==1) {
			ans++;
			if(!tmp) tmp=true,fst=i,snd=i;
			else {
				fst=snd,snd=i;
				REP(j,26) {
					if(s[i]-'a'==j) continue;
					if((cnt2[j]-cnt[j][fst])%2==1&&cnt2[j]!=1) {
						ans++;
					}
				}
			}
		}
	}
	REP(i,26) {
		if(cnt2[i]%2==1&&cnt2[i]!=1&&!flg[i]) ans++;
	}
	cout<<ans<<endl;

	return 0;
}