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 Pascal (FPC 2.6.2) (judge verdict: Wrong Answer)

var s:ansistring;
    num:array[0..200050,'a'..'z']of longint;
    dp:array[0..200050]of longint;
    cannot:boolean;
    i,j,x:longint;
    can:longint;
    n:longint;
begin
  readln(s);
  n:=length(s);
  for i:=1 to n do
  begin
    inc(num[i,s[i]]);
    if i mod 2=1 then can:=1 else can:=0;
    for j:=97 to 122 do
    num[i,char(j)]:=num[i,char(j)]+num[i-1,char(j)];
  end;
  for i:=1 to n do
  begin
    for x:=0 to i-1 do
    begin
      cannot:=false;
      if (i-x) mod 2=1 then can:=1 else can:=0;
      for j:=97 to 122 do
      if (num[i,char(j)]-num[x,char(j)]) mod 2=1 then
      begin
        dec(can);
        if can<0 then
        begin
          cannot:=true;
          break;
        end;
      end;
      if not cannot then
      begin
        dp[i]:=dp[x]+1;
        break;
      end;
    end;
  end;
  writeln(dp[n]);
end.