#include <bits/stdc++.h>
/*|In The Name of Allah|*/
/* Dear GCC compiler:
*
* If my code doesn't compile, this doesn't mean it's wrong.
* It just mean that you don't understand what I want.
* Here's my safety pig, I hope my code compile and get accepted.
* _._ _..._ .-',     _.._(`))
*'-. `     '  /-._.-'    ',/
*   )         \            '.
*  / _    _    |             \
* |  a    a    /              |
* \   .-.                     ;
*  '-('' ).-'       ,'       ;
*     '-;           |      .'
*        \           \    /
*        | 7  .__  _.-\   \
*        | |  |  ``/  /`  /
*       /,_|  |   /,_/   /
*          /,_/      '`-'
*/
using namespace std;
long long power(long long x, long long y){
if (y == 0)
	return 1;
long long ret = power(x , y >> 1) %1000000007 ;
if (y % 2 == 0)
	return (ret * ret) % 1000000007;
else
	return (x * (ret * ret % 1000000007)) % 1000000007;
}

pair<int, int> norm(pair<int, int> p)
{
	if(p.first < 0)
	{
		p.first *= -1;
		p.second *= -1;
	}
	else if (p.first == 0 && p.second < 0)
	{
		p.second *= -1;
	}
	int d = __gcd(abs(p.first), abs(p.second));
	p.first /= d;
	p.second /= d;
	return p;
}

/*

int parent[N], sz[N];
int n, m, components;

void init(){
  iota(parent, parent+n, 0);
  fill(sz, sz+n, 1);
  components = n;
}

int find(int u){
  if(parent[u] == u)  return u;
  return parent[u] = find(parent[u]);
}

void join(int u, int v){
  u = find(u), v = find(v);
  if(u == v)  return;
  if(sz[u] < sz[v]) swap(u, v);
  parent[v] = u;
  sz[u] += sz[v];
  --components;
}
int dst[N];
int vis[N];
int parent[N];
vector<vector<pair<int,ll>>>v(N);
void dikstra(){
	set<pair<ll,ll>>s;
	clr(dst,oo);
	clr(vis,0);
	clr(parent,-1);
	dst[1]=0;
	parent[1]=-1;
	s.insert( mp(0,1 ) );
	while(!s.empty()){

		pair<ll,ll> x=*s.begin();
		s.erase(s.begin());
		ll u=x.sc;
		ll d=x.fr;
		if (vis[u])continue;
		vis[u]=1;
		for (int i=0;i<v[u].size();i++){
			ll to=v[u][i].fr;
			ll cost=v[u][i].sc;
			if (vis[to]||d+cost>=dst[to])continue;
			dst[to]=d+cost;
			parent[to]=u;
			s.insert(mp(d+cost,to));
		}
	}


return;
}
void primeFactors(int n)
{
	while (n % 2 == 0){
		n = n/2;
	}

	for (int i = 3; i <= sqrt(n); i = i + 2)
	{
		while (n % i == 0){
			cout << i << " ";
			n = n/i;
		}
	}
	if (n > 2){


	}
}

void Theaf(){
isPrime[0]=1;
isPrime[1]=1;
for (int i=2;i*i<=N;i++){
	if (isPrime[i]==1)continue;
	for (int cnt=2;i*cnt<=N;cnt++)isPrime[ cnt*i ]=1;
}

}
void dfs_EdgeClassification(int node)
{
	start[node] = timer++;

	rep(i, adj[node])
	{
		int child = adj[node][i];
		if (start[child] == -1)	// Not visited Before. Treed Edge
			dfs_EdgeClassification(child);
		else {
			if(finish[child] == -1)// then this is ancestor that called us and waiting us to finish. Then Cycle. Back Edge
				anyCycle = 1;
			else if(start[node] < start[child])	// then you are my descendant
				;	// Forward Edge
			else
				;	// Cross Edge
		}
	}

	finish[node] = timer++;
}

vectorll factorization(ll n) 	 max n is 1e12
{	// O( sqrt(n) )

	vectorll primes;

	for (ll i = 2; i  i * i <= n; ++i)	 Improve start by i = 3.
		while (n % i == 0)
			primes.push_back(i), n /= i;	 Get every prime inside n.	n  i^j	is a new number

	 Two ways to prove that n will be prime number if n  1
	 1) Think in n  pi^x as a NEW sub-problem. If we went to its sqrt and % applied, then we are facing a prime number
	 2) Prove that, for any n either all its primes before sqrt(n), or only the biggest prime after sqrt(n).
	if (n  > 1)
		primes.push_back(n);

	return primes;
}
int mod(string num, int a)
{
	// Initialize result
	int res = 0;

	// One by one process all digits of 'num'
	for (int i = 0; i < num.length(); i++)
		 res = (res*10 + (int)num[i] - '0') %a;

	return res;
}



struct uf{
	static const int MAXN=105;
	int par[MAXN];
	int size[MAXN];
	void init(){
		memset(par,-1,sizeof(par));
		REP(i,MAXN) size[i]=1;
	}
	int root(int a){
		if(par[a]==-1) return a;
		return par[a]=root(par[a]);
	}
	void unite(int a,int b){
		a=root(a);b=root(b);
		if(a==b) return;
		if(size[a]<size[b]) swap(a,b);

		par[b]=a;
		size[a]+=size[b];
	}
	bool same(int a,int b){
		return root(a)==root(b);
	}
};
uf u_[105];
int mood(string num, int a)
{
	// Initialize result
	int res = 0;

	// One by one process all digits of 'num'
	for (int i = 0; i < (int)num.length(); i++)
		 res = (res*10 + (int)num[i] - '0') %a;

	return res;
}


*/
const double Eps =(1e-7);
int dcmp(double x,double y){ return fabs(x-y)<=Eps? 0: x<y ?-1:1; }
typedef long long  ll;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector<string> vs;
typedef vector<pair<int,int> > vii;
typedef map<int, int> mii;
typedef map<string, int> msi;
typedef pair<int, int> pii;
#define mii map <int, int>
#define mll map <ll, ll>
#define pii pair <int, int>
#define pll pair <ll, ll>
#define vi vector <int>
#define vd vector <double>
#define vll vector <ll>
#define sll set <ll>
#define spii set <pii>
#define vpii vector <pair <int,int> >
#define vpll vector <pair <long long,long long> >
#define vvi vector <vector <int> >
#define vvpii vector <vector <pii > >
#define sfi1(v) scanf("%d",&v)
#define sfi2(v1,v2) scanf("%d %d",&v1,&v2)
#define sfi3(v1,v2,v3) scanf("%d %d %d",&v1,&v2,&v3)
#define sfll1(v) scanf("%I64d",&v);
#define sfll2(v1,v2) scanf("%I64d %I64d",&v1,&v2)
#define sfll3(v1,v2,v3) scanf("%I64d %I64d %I64d",&v1,&v2,&v3)
#define min3(a,b,c)     min(a,min(b,c))
#define max3(a,b,c)     max(a,max(b,c))
#define min4(a,b,c,d)   min(a,min(b,min(c,d)))
#define max4(a,b,c,d)   max(a,max(b,max(c,d)))
#define el	'\n'
#define lp(i,n) for(int i = 0;i < int(n);i++)
#define lp1(i,n) for(int i = 1;i <= int(n);i++)
#define rlp(i, n) for (int i = (n) - 1; i >= 0; i--)
#define rng(i,a,b) for(int i = int(a);i <= int(b);i++)
#define mp(x,y)  	make_pair((x),(y))
#define pb(x)    	push_back((x))
#define all(v)  	((v).begin()),((v).end())
#define sz(x)  		((int) (x).size())
#define bits(a) 	__builtin_popcountll(a)
#define fr first
#define sc second
#define pii pair<int, int>
#define sortva(v) sort(all(v))
#define sortvd(v) sort(v.rbegin(),v.rend())
#define sortaa(a,n) sort(a,a+n)
#define sortad(a,n) sort(a,a+n),reverse(a,a+n)
#define SS stringstream
#define ll long long
#define clr(x, y) memset(x, y, sizeof(x))
#define foreach(it,x) for(__typeof__((x).begin()) it = (x).begin(); it != (x).end(); ++it)
#define read freopen("input.in", "rt", stdin);
#define write freopen("output.in", "wt", stdout);
#define fastIO cout << fixed << setprecision(0), ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
int dx[] = { 0, 0, 1, -1, 1, -1, 1, -1 };
int dy[] = { 1, -1, 0, 0, -1, 1, 1, -1 };
int kx[] = { 2, 2, 1, -1, -2, -2, 1, -1 };
int ky[] = { 1, -1, 2, 2, 1, -1, -2, -2 };
ll gcd(ll a, ll b){if (a == 0)return b;return gcd(b % a, a);}
ll lcm(ll a, ll b) { return (a / gcd(a, b)) * b; }
int const N =5e6+9, mod = 100000000 , oo = 2e9+7;
//------------------------------------------------------------------------------------

string s;
map<char ,int>m;int f=0;
int main()
{

cin>>s;
lp(i,sz(s)){
m[ s[i] ]++;
}
foreach(it,m){
if ((*it).sc%2){
	cout<< "No";
	f=1;
	break;
}
}

if (!f)cout<< "Yes";

















	/*
	*/

    return 0;
}


Measured memory_bytes: 256