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"
#define rep(i,n) for(int i=0;i<n;i++)
#define ALL(v) (v).begin(),(v).end()
typedef long long LL;
const int INF = 1 << 25;
const LL MOD = 1000000007LL;
using namespace std;
bool odd[200001][26]; 
int dp[200001];
int main() {
	string s;
	cin >> s;
	rep(i, s.size()) {
		rep(j, 26) odd[i + 1][j] = odd[i][j];
		odd[i + 1][s[i] - 'a'] ^= 1;
	}
	fill((int*)dp, (int*)(dp + 200001), 1 << 30);
	dp[0] = 0;
	dp[1] = 1;
	for (int i = 2; i <= s.size(); i++) {
		for (int j = 1; j <= i; j++) {
			int sum = 0;
			rep(k, 26) {
				if (odd[i][k] ^ odd[i - j][k]) sum++;
			}
			if (sum <= 1) dp[i] = min(dp[i], dp[i - j] + 1);
		}
	}
	cout << dp[s.size()] << endl;
}