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 <vector>
#include <list>
#include <map>
#include <set>
#include <queue>
#include <deque>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <cstring>
#include <unordered_map>
#include <unordered_set>
#include <cassert>
using namespace std;
typedef pair<int, int> PP;
typedef long long LL;
#define pb push_back
#define fr first
#define sc second
#define bitcnt __builtin_popcount
#define all(x) x.begin(), x.end()
inline LL ri() {LL x; scanf("%lld", &x); return x;}
#define rep2(i, n, ...) for (int i = 0; i < (n); i ++) 
#define rep3(i, a, b, ...) for (int i = (a); i < (b); i ++)
#define GET_MACRO(_1, _2, _3, NAME, ...) NAME
#define rep(...) GET_MACRO(__VA_ARGS__, rep3, rep2)(__VA_ARGS__)
#define drep2(i, n, ...) for (int i = (n) - 1; i >= 0; i --)
#define drep3(i, a, b) for (int i = (a) - 1; i >= (b); i --)
#define drep(...) GET_MACRO(__VA_ARGS__, drep3, drep2)(__VA_ARGS__)
template<typename T>inline bool smax(T&a, T b){if(a<b){a=b;return true;}return false;} 
template<typename T>inline bool smin(T&a, T b){if(a>b){a=b;return true;}return false;} 
/*}}}*/


const int maxn = 200010;
int c[1 << 26], t[1 << 26];
int n;
int main() {
  string s;
  ios::sync_with_stdio(false);
  cin >> s; 
  n = s.length();
  rep(i, 1<<26) c[i] = n;
  c[0] = 0; t[0] = 0;
  int mask = 0;
  rep(i, n) {
    mask ^= 1 << (s[i] - 'a'); 
    if (c[mask] < n) continue;
    rep(j, 26) {
      int tmask = mask ^ (1 << j);
      if (c[tmask] < n) {
        smin(c[mask], c[tmask] + 1);
      }
    }
    t[mask] = i;
  }


  int res = n;
  if (t[mask] != n - 1) {
    res = c[mask] + 1;
    rep(j, 26) {
      int tmask = mask ^ (1 << j);
      smin(res, c[tmask] + 1);
    }
    smin(res, c[mask] + 1);
  } else {
    res = c[mask];
  }

  cout << res << endl;

  return 0;
}