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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)


import java.util.*;
import java.io.*;
import java.lang.reflect.Array;

public class Main {
	public static void main(String[] args) {
		FastReader r = new FastReader();
		HashMap<Character, Integer> map = new HashMap<>();
		for(char c : r.str().toCharArray())
		{
			if(Character.isLowerCase(c)) {
				map.put(c, map.getOrDefault(c, 0)+1);
			}
		}
		for(Map.Entry<Character,Integer> ent : map.entrySet())
		{
			if((ent.getValue()%2)==1)
			{
				print("No");
				return;
			}
		}
		print("Yes");
	}

	public static void rolling(int[] arr) {
		if (arr.length < 2)
			return;
		for (int i = 1; i < arr.length; i++)
			arr[i] += arr[i - 1];
	}
	
	public static void rolling(long[] arr) {
		if (arr.length < 2)
			return;
		for (int i = 1; i < arr.length; i++)
			arr[i] += arr[i - 1];
	}

	public static void reverse(int[] arr) {
		for (int i = 0; i < arr.length / 2; i++) {
			int temp = arr[i];
			arr[i] = arr[arr.length - 1 - i];
			arr[arr.length - 1 - i] = temp;
		}
	}

	public static void print(String s) {
		System.out.println(s);
	}

	public static void print(long s) {
		System.out.println(s);
	}

	public static void print(int s) {
		System.out.println(s);
	}

	static class FastReader {
		BufferedReader br;
		StringTokenizer st;

		public FastReader() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}

		int ri() {
			return Integer.parseInt(str());
		}

		long rl() {
			return Long.parseLong(str());
		}

		double rd() {
			return Double.parseDouble(str());
		}

		String str() {
			String str = "";
			try {
				str = br.readLine();
			} catch (IOException e) {
				e.printStackTrace();
			}
			return str;
		}

		String[] rsa() {
			return str().split(" ");
		}

		int[] ria() {
			String[] s = rsa();
			int[] arr = new int[s.length];
			for (int i = 0; i < arr.length; i++)
				arr[i] = Integer.parseInt(s[i]);
			return arr;
		}

		long[] rla() {
			String[] s = rsa();
			long[] arr = new long[s.length];
			for (int i = 0; i < arr.length; i++)
				arr[i] = Long.parseLong(s[i]);
			return arr;
		}

		char[] rca() {
			String s = str();
			char[] arr = new char[s.length()];
			for (int i = 0; i < arr.length; i++)
				arr[i] = s.charAt(i);
			return arr;
		}
		
		String[] rsaml(int m) {
			String[] arr = new String[m];
			for (int i = 0; i < arr.length; i++)
				arr[i] = str();
			return arr;
		}

		int[] riaml(int m) {
			int[] arr = new int[m];
			for (int i = 0; i < arr.length; i++)
				arr[i] = Integer.parseInt(str());
			return arr;
		}

		long[] rlaml(int m) {
			long[] arr = new long[m];
			for (int i = 0; i < arr.length; i++)
				arr[i] = Long.parseLong(str());
			return arr;
		}
	}
}

class Node {
	public int val;
	public boolean visited;
	public Node(int v) { val=v; }
}