Problem:
Score : 200 points

Problem Statement

Let w be a string consisting of lowercase letters.
We will call w beautiful if the following condition is satisfied:

Each lowercase letter of the English alphabet occurs even number of times in w.

You are given the string w. Determine if w is beautiful.

Constraints

1 \leq |w| \leq 100

w consists of lowercase letters (a-z).

Input

The input is given from Standard Input in the following format:

w

Output

Print Yes if w is beautiful. Print No otherwise.

Sample Input 1

abaccaba

Sample Output 1

Yes

a occurs four times, b occurs twice, c occurs twice and the other letters occur zero times.

Sample Input 2

hthth

Sample Output 2

No

Solution in C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <vector>
#include <iterator>
#include <algorithm>
#include <string>
#include <map>
#include <list>

using namespace std;

int main() {
  string s;
  cin >> s;
  int a[]={};

  bool can = true;
  int N = s.size();
  for (int i=0; i <= 24; i++){
    a[i] = 0;
  }
  for (int i=0; i < N; i++){
    if(s[i] == 'a'){
      a[0]++;
    }
    if(s[i] == 'b'){
      a[1]++;
    }
    if(s[i] == 'c'){
      a[2]++;
    }
    if(s[i] == 'd'){
      a[3]++;
    }
    if(s[i] == 'e'){
      a[4]++;
    }
    if(s[i] == 'f'){
      a[5]++;
    }
    if(s[i] == 'g'){
      a[6]++;
    }
    if(s[i] == 'h'){
      a[7]++;
    }
    if(s[i] == 'i'){
      a[8]++;
    }
    if(s[i] == 'j'){
      a[9]++;
    }
    if(s[i] == 'k'){
      a[10]++;
    }
    if(s[i] == 'l'){
      a[11]++;
    }
    if(s[i] == 'm'){
      a[12]++;
    }
    if(s[i] == 'n'){
      a[13]++;
    }
    if(s[i] == 'o'){
      a[14]++;
    }
    if(s[i] == 'p'){
      a[15]++;
    }
    if(s[i] == 'q'){
      a[16]++;
    }
    if(s[i] == 'r'){
      a[17]++;
    }
    if(s[i] == 's'){
      a[18]++;
    }
    if(s[i] == 't'){
      a[19]++;
    }
    if(s[i] == 'u'){
      a[20]++;
    }
    if(s[i] == 'v'){
      a[21]++;
    }
    if(s[i] == 'w'){
      a[22]++;
    }
    if(s[i] == 'y'){
      a[23]++;
    }
    if(s[i] == 'z'){
      a[24]++;
    }
  }
  for (int i=0; i <= 24; i++){
    if(a[i] % 2 == 0){
      can = true;
    }else
    {
      can = false;
    }
  }
  if (can == true){
    cout << "YES" << endl;
  }else{
    cout << "NO" << endl;
  }
}