Problem: An integer sequence is called beautiful if the difference between any two consecutive numbers is equal to 1. More formally, a sequence s_1, s_2, …, s_{n} is beautiful if |s_i - s_{i+1}| = 1 for all 1 ≤ i ≤ n - 1.

Trans has a numbers 0, b numbers 1, c numbers 2 and d numbers 3. He wants to construct a beautiful sequence using all of these a + b + c + d numbers.

However, it turns out to be a non-trivial task, and Trans was not able to do it. Could you please help Trans?

Input

The only input line contains four non-negative integers a, b, c and d (0 < a+b+c+d ≤ 10^5).

Output

If it is impossible to construct a beautiful sequence satisfying the above constraints, print "NO" (without quotes) in one line.

Otherwise, print "YES" (without quotes) in the first line. Then in the second line print a + b + c + d integers, separated by spaces — a beautiful sequence. There should be a numbers equal to 0, b numbers equal to 1, c numbers equal to 2 and d numbers equal to 3.

If there are multiple answers, you can print any of them.

Examples

Input


2 2 2 1


Output


YES
0 1 0 1 2 3 2


Input


1 2 3 4


Output


NO


Input


2 2 2 3


Output


NO

Note

In the first test, it is easy to see, that the sequence is beautiful because the difference between any two consecutive numbers is equal to 1. Also, there are exactly two numbers, equal to 0, 1, 2 and exactly one number, equal to 3.

It can be proved, that it is impossible to construct beautiful sequences in the second and third tests.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <typename T>
using pq = priority_queue<T, vector<T>, greater<T>>;
using ll = long long;
ll INF = LLONG_MAX / 4000 - 100;
ll mod = 1e9 + 7;
long long dx[] = {-1, 0, 1, 0}, dy[] = {0, 1, 0, -1};
vector<ll> prime;
double pi = 3.141592653689793238;
class fact {
 public:
  long long fmod = 1e9 + 7;
  vector<long long> fac, finv, inv;
  fact(long long n, long long Mod = 1e9 + 7) {
    fmod = Mod;
    fac = vector<long long>(n + 1, 0);
    finv = vector<long long>(n + 1, 0);
    inv = vector<long long>(n + 1, 0);
    fac[0] = 1;
    for (long long i = 1; i < n + 1; i++) fac[i] = fac[i - 1] * i % fmod;
    for (long long i = 0; i < n + 1; i++) finv[i] = fact::POW(fac[i], fmod - 2);
    for (long long i = 0; i < n + 1; i++) inv[i] = POW(i, fmod - 2);
  }
  ll nCr(ll n, ll r) {
    if (n < r) return 0;
    return (fac[n] * finv[r] % fmod) * finv[n - r] % fmod;
  }
  ll POW(ll a, ll b) {
    ll c = 1;
    while (b > 0) {
      if (b & 1) {
        c = a * c % fmod;
      }
      a = a * a % fmod;
      b >>= 1;
    }
    return c;
  }
  inline long long operator[](long long i) { return fac[i]; }
};
void DEBUG(vector<long long> a) {
  for (long long i = 0; i < a.size(); i++) cout << a[i] << " ";
  cout << endl;
}
void EMP(long long x) { cout << "!!!" << x << "!!!" << endl; }
ll GCD(ll a, ll b) {
  ll c;
  while (b != 0) {
    c = a % b;
    a = b;
    b = c;
  }
  return a;
}
ll LCM(ll a, ll b) { return (a / GCD(a, b)) * (b / GCD(a, b)) * GCD(a, b); }
ll POW(ll a, ll b) {
  ll c = 1;
  while (b > 0) {
    if (b & 1) {
      c = a * c % mod;
    }
    a = a * a % mod;
    b >>= 1;
  }
  return c;
}
void PRI(ll n) {
  bool a[n + 1LL];
  for (long long i = 0; i < n + 1LL; i++) {
    a[i] = 1LL;
  }
  for (long long i = 2; i < n + 1LL; i++) {
    if (a[i]) {
      prime.push_back(i);
      ll b = i;
      while (b <= n) {
        a[b] = 0;
        b += i;
      }
    }
  }
}
template <typename T>
T chmin(T& a, T b) {
  if (a > b) a = b;
  return a;
}
template <typename T>
T chmax(T& a, T b) {
  if (a < b) a = b;
  return b;
}
bool isSqrt(ll a) { return pow(sqrt(a), 2) == a ? 1 : 0; }
void YesNo(bool a) {
  if (a)
    cout << "Yes";
  else
    cout << "No";
  cout << endl;
}
void yesno(bool a) {
  if (a)
    cout << "yes";
  else
    cout << "no";
  cout << endl;
}
void YESNO(bool a) {
  if (a)
    cout << "YES";
  else
    cout << "NO";
  cout << endl;
}
double dis(long long x1, long long x2, long long y1, long long y2) {
  return sqrt((double)abs(x1 - x2) * (double)abs(x1 - x2) +
              (double)abs(y1 - y2) * (double)abs(y1 - y2));
}
long long ceili(long long x, long long y) {
  if (x % y == 0)
    return x / y;
  else
    return x / y + 1;
}
long long extGCD(long long a, long long b, long long& x, long long& y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  }
  long long d = extGCD(b, a % b, y, x);
  y -= a / b * x;
  return d;
}
long long a[4];
vector<long long> ans(100005);
bool func(long long f, long long del) {
  ans = vector<long long>(100005);
  long long cnt[4];
  for (long long(i) = (0); (i) < (4); (i)++) cnt[i] = a[i];
  if (cnt[f] == 0) return false;
  cnt[f]--;
  ans[0] = f;
  long long now = f + del;
  for (long long(i) = (1); (i) < (a[0] + a[1] + a[2] + a[3]); (i)++) {
    if (now < 0 || 3 < now) return false;
    if (cnt[now] == 0) return false;
    cnt[now]--;
    ans[i] = now;
    if (cnt[now - del] > 0) del *= -1;
    now += del;
  }
}
void solve() {
  for (long long(i) = (0); (i) < (4); (i)++) cin >> a[i];
  if (func(0, 1)) {
    cout << "YES" << endl;
    for (long long(i) = (0); (i) < (a[0] + a[1] + a[2] + a[3]); (i)++)
      cout << ans[i] << " ";
    cout << endl;
    return;
  }
  if (func(1, -1)) {
    cout << "YES" << endl;
    for (long long(i) = (0); (i) < (a[0] + a[1] + a[2] + a[3]); (i)++)
      cout << ans[i] << " ";
    cout << endl;
    return;
  }
  if (func(1, 1)) {
    cout << "YES" << endl;
    for (long long(i) = (0); (i) < (a[0] + a[1] + a[2] + a[3]); (i)++)
      cout << ans[i] << " ";
    cout << endl;
    return;
  }
  if (func(2, -1)) {
    cout << "YES" << endl;
    for (long long(i) = (0); (i) < (a[0] + a[1] + a[2] + a[3]); (i)++)
      cout << ans[i] << " ";
    cout << endl;
    return;
  }
  if (func(2, 1)) {
    cout << "YES" << endl;
    for (long long(i) = (0); (i) < (a[0] + a[1] + a[2] + a[3]); (i)++)
      cout << ans[i] << " ";
    cout << endl;
    return;
  }
  if (func(3, -1)) {
    cout << "YES" << endl;
    for (long long(i) = (0); (i) < (a[0] + a[1] + a[2] + a[3]); (i)++)
      cout << ans[i] << " ";
    cout << endl;
    return;
  }
  cout << "NO" << endl;
  return;
}
signed main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  solve();
}