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 <cstdio>
#include <map>
#include <vector>
using namespace std;
char s[200010];
int dp[1<<26], q[200010], qs;
map<int, vector<int> > e;
bool can[1<<26];
int main() {
	int l;
	scanf("%s", s);
	for (l = 0; s[l]; l++) ;
	int b = 0;
	for (int i = 0; i < l; i++) {
		int x = s[i] - 'a';
		b ^= 1 << x;
	}
	int tb = b;
	can[b] = 1;
	for (int i = 0; i < 1 << 26; i++) {
		dp[i] = -1;
	}
	for (int i = l - 1; i >= 0; i--) {
		int x = s[i] - 'a';
		b ^= 1 << x;
		can[b] = 1;
		for (int j = 0; j < 26; j++) {
			int y = b ^ (1 << j);
			if (can[y]) {
				e[b].push_back(y);
			}
		}
	}
	dp[0] = 0;
	qs++;
	int p = 0;
	while (p < qs) {
		int x = q[p++];
		vector<int> &to = e[x];
		for (int i = 0; i < to.size(); i++) {
			int y = to[i];
			if (dp[y] == -1) {
				dp[y] = dp[x] + 1;
				q[qs++] = y;
			}
		}
	}
	printf("%d\n", dp[tb] == 0 ? 1 : dp[tb]);
	return 0;
}