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: Runtime Error)

#include "bits/stdc++.h"

#define REP(i,n) for(ll i=0;i<ll(n);++i)
#define RREP(i,n) for(ll i=ll(n)-1;i>=0;--i)
#define FOR(i,m,n) for(ll i=m;i<ll(n);++i)
#define RFOR(i,m,n) for(ll i=ll(n)-1;i>=ll(m);--i)
#define ALL(v) (v).begin(),(v).end()
#define UNIQUE(v) v.erase(unique(ALL(v)),v.end());
#define DUMP(v) REP(aa, (v).size()) { cout << v[aa]; if (aa != v.size() - 1)cout << " "; else cout << endl; }
#define INF 1000000001ll
#define MOD 1000000007ll
#define EPS 1e-9

const int dx[8] = { 1,1,0,-1,-1,-1,0,1 };
const int dy[8] = { 0,1,1,1,0,-1,-1,-1 };


using namespace std;
typedef long long ll;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<vi> vvi;
typedef vector<vl> vvl;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
ll max(ll a, int b) { return max(a, ll(b)); }
ll max(int a, ll b) { return max(ll(a), b); }
ll min(ll a, int b) { return min(a, ll(b)); }
ll min(int a, ll b) { return min(ll(a), b); }
///(´・ω・`)(´・ω・`)(´・ω・`)(´・ω・`)(´・ω・`)(´・ω・`)///
string s;
int n;
int dfs(vi &dp, vi &pos, vi &bit, int i) {
	if (i == n)return 0;
	if (dp[i] != -1)return dp[i];
	//cout << i << endl;
	int ret = INF;
	int b = bit[n]^bit[i];
	int popcount = 0;
	REP(j, 26) {
		if ((1 << j)&b)popcount++;
	}
	if (popcount <= 1)return dp[i] = 1;

	b = bit[i];
	if (pos[b] >= i) {
		assert((pos[b] - i) % 2 == 0);
		ret = min(ret, 1 + dfs(dp, pos, bit, pos[b] + 1));
	}
	REP(k, 26) {
		int c = b ^ (1 << k);
		
		if (pos[c] >= i) {
			assert((pos[c] - i) % 2 == 1);
			ret = min(ret, 1 + dfs(dp, pos, bit, pos[c] + 1));
		}
	}
	return dp[i] = ret;

}

int main() {
	cin.tie(0);
	ios::sync_with_stdio(false);

	cin >> s;
	n = s.size();
	vi bit(n + 1);

	vi pos(1 << 26,-1);
	bit[0] = 0;
	REP(i, n) {
		bit[i + 1] = bit[i] ^ (1 << (s[i] - 'a'));
		pos[bit[i + 1]]=i;
	
	}
	vi dp(n, -1);
	dfs(dp, pos, bit, 0);
	cout << dp[0] << endl;
}