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;
inline void printArr(vector<long long> &arr) {
  for (long long i : arr) cout << i << " ";
  cout << '\n';
}
inline void printArr(vector<long long> &arr, long long n) {
  for (long long i = 0; i < n; ++i) cout << i << " ";
  cout << '\n';
}
inline void printArr(vector<long long> &arr, long long start, long long n) {
  for (long long i = start; i < start + n; ++i) cout << i << " ";
  cout << '\n';
}
const long long M = 1e9 + 7;
vector<long long> topologicalOrder(long long V, vector<long long> adj[]) {
  vector<long long> inDegree(V, 0);
  for (long long i = 0; i < V; ++i) {
    for (long long j = 0; j < adj[i].size(); ++j) ++inDegree[adj[i][j]];
  }
  queue<long long> q;
  for (long long i = 0; i < V; ++i) {
    if (inDegree[i] == 0) q.push(i);
  }
  vector<long long> topOrder;
  while (!q.empty()) {
    long long front = q.front();
    q.pop();
    topOrder.push_back(front);
    for (long long i = 0; i < adj[front].size(); ++i) {
      --inDegree[adj[front][i]];
      if (inDegree[adj[front][i]] == 0) q.push(adj[front][i]);
    }
  }
  return topOrder;
}
long long powMod(long long a, long long b, long long m) {
  long long x = 1;
  while (b > 0) {
    if (b & 1) x = (x * a) % m;
    a = (a * a) % m;
    b >>= 1;
  }
  return x;
}
long long mulMod(long long a, long long b, long long m) {
  long long res = 0;
  a = a % m;
  while (b > 0) {
    if (b % 2 == 1) res = (res + a) % m;
    a = (a * 2) % m;
    b /= 2;
  }
  return res % m;
}
long long nCr(long long n, long long r, long long m) {
  if (n < r) return 0;
  long long res = 1, rem = 1;
  for (long long i = n - r + 1; i <= n; i++) res = (res * i) % m;
  for (long long i = 2; i <= r; i++) rem = (rem * i) % m;
  return (res * powMod(rem, m - 2, m)) % m;
}
long long modInverse(long long a, long long m) {
  a = a % m;
  for (long long x = 1; x < m; ++x) {
    if ((a * x) % m == 1) return x;
  }
  return -1;
}
long long gcdExtended(long long a, long long b, long long *x, long long *y) {
  if (b == 0) {
    *x = 1, *y = 0;
    return a;
  }
  long long x1, y1;
  long long gcd = gcdExtended(b, a % b, &x1, &y1);
  *x = y1, *y = x1 - (floor(a / b) * y1);
  return gcd;
}
long long factorial(long long n) {
  if (n == 1) return 1;
  return sqrt(2 * 3.141592653589793238461 * n) *
         pow((n / 2.71828182845904523536), n);
}
vector<bool> sieve(long long n) {
  long long sqrtn = sqrt(n);
  vector<bool> is_prime(n + 1, true);
  for (long long i = 3; i <= sqrtn; i += 2) {
    if (is_prime[i]) {
      long long increment = i + i;
      for (long long j = i * i; j <= n; j += increment) is_prime[j] = false;
    }
  }
  return is_prime;
}
vector<pair<long long, long long>> primeFactors(long long n) {
  vector<pair<long long, long long>> ans;
  for (long long i = 2; i * i <= n; ++i) {
    if (n % i == 0) {
      long long cnt = 0;
      while (n % i == 0) n /= i, cnt++;
      if (cnt) ans.push_back({i, cnt});
    }
  }
  if (n > 1) ans.push_back({n, 1});
  return ans;
}
void factors(vector<pair<long long, long long>> primeFactors,
             vector<long long> &ans, long long cur = 0, long long curRes = 1) {
  if (cur == primeFactors.size()) {
    ans.push_back(curRes);
    return;
  }
  for (long long i = 0; i <= primeFactors[cur].second; ++i) {
    factors(primeFactors, ans, cur + 1, curRes);
    curRes *= primeFactors[cur].first;
  }
}
vector<vector<long long>> matMul(vector<vector<long long>> A,
                                 vector<vector<long long>> B) {
  vector<vector<long long>> C(A.size(), vector<long long>(B[0].size()));
  for (long long i = 0; i < A.size(); ++i) {
    for (long long j = 0; j < B[0].size(); ++j) {
      C[i][j] = 0;
      for (long long k = 0; k < B.size(); ++k)
        C[i][j] = (C[i][j] + ((A[i][k] * B[k][j]) % M)) % M;
    }
  }
  return C;
}
vector<vector<long long>> matPow(vector<vector<long long>> A, long long p) {
  if (p == 1) return A;
  if (p & 1)
    return matMul(A, matPow(A, p - 1));
  else {
    vector<vector<long long>> C = matPow(A, p / 2);
    return matMul(C, C);
  }
}
vector<long long> zFunction(string s) {
  long long n = s.size(), L = 0, R = 0;
  vector<long long> z(n);
  for (long long i = 1; i < n; ++i) {
    if (i > R) {
      L = R = i;
      while (R < n && s[R - L] == s[R]) R++;
      z[i] = R - L;
      R--;
    } else {
      long long k = i - L;
      if (z[k] < R - i + 1)
        z[i] = z[k];
      else {
        L = i;
        while (R < n && s[R - L] == s[R]) R++;
        z[i] = R - L;
        R--;
      }
    }
  }
  return z;
}
vector<long long> patterMatch(string text, string pattern) {
  string key = pattern + '$' + text;
  vector<long long> z = zFunction(key);
  vector<long long> res;
  long long m = pattern.size();
  for (long long i = m + 1; i < key.size(); ++i)
    if (z[i] == m) res.push_back(i - m - 1);
  return res;
}
vector<long long> kaden(vector<long long> &arr) {
  vector<long long> kaden(arr.size());
  kaden[0] = arr[0];
  for (long long i = 1; i < arr.size(); ++i)
    kaden[i] = max(arr[i], arr[i] + kaden[i - 1]);
  return kaden;
}
long long digitSum(long long x) {
  long long ds = 0;
  while (x) {
    ds += x % 10;
    x /= 10;
  }
  return ds;
}
long long a[1000005], b[1000005];
vector<long long> g[1000005];
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> a[0] >> a[1] >> a[2] >> a[3];
  g[0].push_back(1);
  g[1].push_back(-1);
  g[1].push_back(1);
  g[2].push_back(1);
  g[2].push_back(-1);
  g[3].push_back(-1);
  for (long long i = 0; i <= 3; ++i) {
    if (a[i] == 0) continue;
    for (long long j = 0; j <= 3; ++j) b[j] = a[j];
    long long j = i;
    vector<long long> ans;
    b[j]--;
    ans.push_back(j);
    while (true) {
      bool ok = false;
      for (long long s = 0; s < g[j].size(); ++s) {
        long long x = g[j][s];
        if (b[j + x] > 0) {
          b[j + x]--;
          ans.push_back(j + x);
          j += x;
          ok = true;
          break;
        }
      }
      if (!ok) break;
    }
    bool ok = true;
    for (long long j = 0; j <= 3; ++j)
      if (b[j] != 0) ok = false;
    if (ok) {
      cout << "YES\n";
      for (long long s = 0; s < ans.size(); ++s) cout << ans[s] << " ";
      return 0;
    }
  }
  cout << "NO\n";
  return 0;
}