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;
typedef long long ll;

int dp[200010];

int main(){
	cin.tie(0);
	ios::sync_with_stdio(false);
#ifdef LOCAL
	std::ifstream in("in");
	std::cin.rdbuf(in.rdbuf());
#endif

	string s;
	cin >> s;
	int N = s.size();
	int bit = 0;
	map<int, int> m;
	m[0] = 0;
	for(int i = 0; i < N; i++){
		//cout << i << endl;
		dp[i + 1] = dp[i] + 1;
		int id = s[i] - 'a';
		// 全部余ってない
		int zero = bit;
		if(m.count(zero)){
			dp[i + 1] = min(dp[i + 1], m[zero] + 1);
		}
		// idが余ってて他余ってない
		int one = bit ^ (1 << id);
		if(m.count(one)){
			dp[i + 1] = min(dp[i + 1], m[one] + 1);
		}
		// idが余ってて他1つ余ってる
		for(int j = 0; j < 26; j++){
			if(j == id) continue;
			one = bit ^ (1 << id) ^ (1 << j);
			if(m.count(one)){
				//cout << char('a' + j) << endl;
				dp[i + 1] = min(dp[i + 1], m[one] + 1);
			}
		}

		if(!m.count(bit)){
			m[bit] = dp[i];
		}
		else{
			m[bit] = min(m[bit], dp[i]);
		}
		bit ^= 1 << id;
	}

	cout << dp[N] << endl;
}