#include <bits/stdc++.h>
using namespace std;

// #define int long long
#define rep(i, n) for (long long i = (long long)(0); i < (long long)(n); ++i)
#define reps(i, n) for (long long i = (long long)(1); i <= (long long)(n); ++i)
#define rrep(i, n) for (long long i = ((long long)(n)-1); i >= 0; i--)
#define rreps(i, n) for (long long i = ((long long)(n)); i > 0; i--)
#define irep(i, m, n) for (long long i = (long long)(m); i < (long long)(n); ++i)
#define ireps(i, m, n) for (long long i = (long long)(m); i <= (long long)(n); ++i)
#define irreps(i, m, n) for (long long i = ((long long)(n)-1); i > (long long)(m); ++i)
#define SORT(v, n) sort(v, v + n);
#define REVERSE(v, n) reverse(v, v+n);
#define vsort(v) sort(v.begin(), v.end());
#define all(v) v.begin(), v.end()
#define mp(n, m) make_pair(n, m);
#define cinline(n) getline(cin,n);
#define replace_all(s, b, a) replace(s.begin(),s.end(), b, a);
#define PI (acos(-1))
#define FILL(v, n, x) fill(v, v + n, x);
#define sz(x) (long long)(x.size())

using ll = long long;
using vi = vector<int>;
using vvi = vector<vi>;
using vll = vector<ll>;
using vvll = vector<vll>;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using vs = vector<string>;
using vpll = vector<pair<ll, ll>>;
using vtp = vector<tuple<ll,ll,ll>>;
using vb = vector<bool>;

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }

template<class t> using vc=vector<t>;
template<class t> using vvc=vc<vc<t>>;

const ll INF = 1e9+10;
const ll MOD = 1e9+7;
const ll LINF = 1e18;



signed main()
{
  cin.tie( 0 ); ios::sync_with_stdio( false );
  
  ll n; cin>>n;
  ll sum=0;
  vll a(n-1);
  rep(i,n-1) {
    cin>>a[i];
    sum+=a[i];
  }
  
  
  // 総和を見れば良さそう
  // 木を作れるかという問題
  // n-1辺で全てを網羅
  // a[i]が1つ存在すれば、a[i]+1のノードへは到達可能
  // b[i]をa[i]で埋めて、埋まってないところは直近の右からの距離になる
  
  // a[i]+1==iなるiを埋める
  // それで埋まらなかったiは、小さい値から使っていく
  
  ll ma=n*(n-1)/2, mi=n-1;
  // cout<<ma<<' '<<mi<<endl;
  if(mi<=sum and sum<=ma) {
    cout<<"YES"<<endl;
    
    vll used(n+1);
    vpll unused;
    vll ans(n-1);
    rep(i,n-1){
      if(used[a[i]]==0){
        used[a[i]]=1;
        ans[i]=1;
      }else{
        unused.emplace_back(a[i],i);
      }
    }
    // irep(i,1,n) cout<<used[i]<<' ';cout<<endl;
    // rep(i,n-1) cout<<ans[i]<<' '; cout<<endl;
    // cout<<unused.size()<<endl;
    vsort(unused);
    ll j=0;
    rep(i,n-1){
      if(!ans[i]){
        ans[unused[j].second]=(i+2)-unused[j].first;
        j++;
      }
    }
    rep(i,n-1) cout<<ans[i]<<endl;
    
  }else{
    cout<<"NO"<<endl;
  }
  
}