#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define P pair<ll,ll>
#define FOR(I,A,B) for(ll I = ll(A); I < ll(B); ++I)
#define FORR(I,A,B) for(ll I = ll((B)-1); I >= ll(A); --I)
#define TO(x,t,f) ((x)?(t):(f))
#define SORT(x) (sort(x.begin(),x.end())) // 0 2 2 3 4 5 8 9
#define POSL(x,v) (lower_bound(x.begin(),x.end(),v)-x.begin()) //xi>=v  x is sorted
#define POSU(x,v) (upper_bound(x.begin(),x.end(),v)-x.begin()) //xi>v  x is sorted
#define NUM(x,v) (POSU(x,v)-POSL(x,v))  //x is sorted
#define REV(x) (reverse(x.begin(),x.end())) //reverse
ll gcd(ll a,ll b){if(a%b==0)return b;return gcd(b,a%b);}
ll lcm(ll a,ll b){ll c=gcd(a,b);return ((a/c)*(b/c)*c);}
#define NEXTP(x) next_permutation(x.begin(),x.end())
const ll INF=ll(1e16)+ll(7);
const ll MOD=1000000007LL;
#define out(a) cout<<fixed<<setprecision((a))
//tie(a,b,c) = make_tuple(10,9,87);
#define pop_(a) __builtin_popcount((a))
ll keta(ll a){ll r=0;while(a){a/=10;r++;}return r;}
ll ketawa(ll a){ll r=0;while(a){r+=a%10;a/=10;}return r;}


int main(){

  ll n,x[3],y[]={1000,5000,10000};
  cin >> n >> x[0] >> x[1] >> x[2];
  priority_queue<ll> pq;

  FOR(i,0,n){
    ll a;
    cin >> a;
    pq.push(a);
  }

  while(pq.size() && x[2]){
    if(pq.top() < 10000){
      x[2]--;
      pq.pop();
      continue;
    }
    ll num = pq.top();
    pq.pop();
    ll a = num;
    num /= 10000;
    num = min(num,x[2]); 
    a -= 10000 * num;
    x[2] -= num;
    pq.push(a);
  }
  while(pq.size() && x[1]){
    if(pq.top() < 5000){
      x[1]--;
      pq.pop();
      continue;
    }
    ll num = pq.top();
    pq.pop();
    ll a = num;
    num /= 5000;
    num = min(num,x[1]); 
    a -= 5000 * num;
    x[1] -= num;
    pq.push(a);
  }
  while(pq.size() && x[0]){
    if(pq.top() < 1000){
      x[0]--;
      pq.pop();
      continue;
    }
    ll num = pq.top();
    pq.pop();
    ll a = num;
    num /= 1000;
    num = min(num,x[0]); 
    a -= 1000 * num;
    x[0] -= num;
    pq.push(a);
  }

  if(pq.size() == 0){
    cout << "Yes" << endl;
  }else{
    cout << "No" << endl;
  }


}