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: Accepted)

#include<bits/stdc++.h>
using namespace std;
using ll=long long;
#define fr(i,n) for(unsigned i=0;i<(n);++i)
#define Fr(i,n) for(int i=1;i<=(n);++i)
#define ifr(i,n) for(int i=(n)-1;i>=0;--i)
#define iFr(i,n) for(int i=(n);i>0;--i)
template<class T> void chmin(T&x,T y){if(x>y)x=y;}
template<class T> void chmax(T&x,T y){if(x<y)x=y;}

inline void PUTS(int a){
    static stack<char> C;
    while(a) C.emplace('0'+a%10),a/=10;
    if(C.empty()) putchar_unlocked('0');
    for(;!C.empty();C.pop()) putchar_unlocked(C.top());
    putchar_unlocked('\n');
}

unsigned c,dp[1u<<26u],d[200001],n{};
int main(){
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    istream& in(cin);
    ostream& out(cout);
    d[0]=0;
    while((c=getchar_unlocked()-'a')<26) d[n+1]=d[n]^(1u<<c),++n;
    fill(dp,dp+(1u<<26u),500000);
    dp[d[n]]=0;
    ifr(i,n) fr(j,26) chmin(dp[d[i]],dp[d[i]^(1u<<j)]+1);
    PUTS(max(dp[0],1u));
}