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 <bits/stdc++.h>
#define mp       make_pair
#define pb       push_back
#define all(x)   (x).begin(),(x).end()
#define YES() printf("YES\n")
#define NO() printf("NO\n")
#define Yes() printf("Yes\n")
#define No() printf("No\n")
#define in(x,y,h,w) x >= 0 && x < h && y >= 0 && y < w
using namespace std;

//#define int long long
//typedef    long long          ll;
typedef    vector<bool>       vb;
typedef    vector<int>        vi;
typedef    vector<vb>         vvb;
typedef    vector<vi>         vvi;
typedef    pair<int,int>      P;
 
const int INF=1e+9;
const double EPS=1e-9;
const int MOD=1000000007;
 
const int dx[]={1,0,-1,0},dy[]={0,-1,0,1};

int dp[1 << 26],tmp = 0;

signed main(){
	string s;
	cin >> s;
	for(int i = 0;i < (1 << 26);i++) dp[i] = INF;
	dp[0] = 0;
	for(int i = 0;i < s.length();i++){
		tmp ^= 1 << (s[i] - 'a');
		for(int j = 0;j < 26;j++){
			dp[tmp] = min(dp[tmp],dp[tmp ^ (1 << j)] + 1);
		}
	}
	cout << dp[tmp] << endl;
	return 0;
}