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: Time Limit Exceeded)

#include<bits/stdc++.h>
using namespace std;
using Int = long long;
signed main(){
  string s;
  cin>>s;
  int n=s.size();
  vector<int> v(n+1);
  v[0]=0;
  map<int,vector<int> > m;
  m[0].push_back(0);
  for(int i=0;i<n;i++){
    v[i+1]=v[i]^(1<<(s[i]-'a'));
    m[v[i+1]].push_back(i+1);
  }
  //cout<<bitset<26>(v[n])<<endl;
  vector<int> dp(n+1,1<<28);
  dp[0]=0;
  for(int i=0;i<n;i++){
    int b=v[i];
    //cout<<i<<":"<<dp[i]<<" "<<bitset<26>(b)<<endl;
    dp[i+1]=min(dp[i+1],dp[i]+1);
    dp[m[b].back()]=min(dp[m[b].back()],dp[i]+1);
    for(int j=0;j<26;j++){
      b^=(1<<j);
      //if(m.count(b)) cout<<i<<":"<<bitset<26>(b)<<" "<<m[b].back()<<endl;
      //cout<<v[i]<<" "<<b<<endl;
      if(m.count(b)){
	for(int k:m[b]){
	  dp[k]=min(dp[k],dp[i]+1);
	}
      }
      b^=(1<<j);
    }
  }
  cout<<dp[n]<<endl;
  return 0;
}