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;
long long MOD = 998244353;
long long power(long long x, long long n, long long mod) {
  if (n == 0) {
    return 1;
  }
  long long temp = power(x, n / 2, mod);
  if (n % 2 == 1) {
    return (x * temp * temp) % mod;
  } else {
    return (temp * temp) % mod;
  }
}
long long abs(long long a, long long b) {
  if (a > b) {
    return a - b;
  }
  return b - a;
}
long long fac(long long n) {
  long long sol = 1;
  for (long long i = 1; i <= n; i++) {
    sol *= i;
    sol %= MOD;
  }
  return sol;
}
long long size(long long n) {
  long long sol = 0;
  while (n) {
    n /= 2;
    sol++;
  }
  return sol;
}
long long fromBinary(string n) {
  long long sol = 0;
  for (long long i = 0; i < n.size(); i++) {
    sol += n[i] - '0';
    sol *= 2;
  }
  return sol / 2;
}
vector<long long> num;
set<long long> num2;
vector<long long> deleted;
void check(long long a, long long b, bool deleteOrNo) {
  if (num2.count(abs(a - b))) {
    check(abs(a - b), a, true);
  } else if (deleteOrNo) {
    num2.erase(a);
    deleted.push_back(a);
  }
}
bool thinksHeIsBetter(long long a, long long b) {
  while (1) {
    if (a % 2 == 1 && b % 2 == 0) {
      return true;
    } else {
      a /= 2;
      b /= 2;
    }
    if (a == 0 && b == 0) {
      return false;
    }
  }
}
long long k;
vector<long long> primes;
pair<long long, long long> makeTwin(long long n) {
  long long twin = 1;
  long long exp;
  long long betterN = 1;
  for (long long i = 0; i <= primes.size() - 1; i++) {
    if (k >= 4 && i > 16 && n > 1) {
      return make_pair(-1, -1);
    }
    if (k >= 3 && i > 70 && n > 1) {
      return make_pair(-1, -1);
    }
    if (i > 70) {
      break;
    }
    if (primes[i] != n) {
      if (primes[i] * primes[i] > n) {
        continue;
      }
    }
    exp = 0;
    while (n % primes[i] == 0) {
      exp++;
      n /= primes[i];
    }
    if (exp == 0) {
      continue;
    }
    exp %= k;
    betterN *= power(primes[i], exp, 1000000000000000);
    if (exp == 0) {
      exp = k;
    }
    twin *= power(primes[i], k - exp, 1000000000000000);
    if (twin > 100000) {
      return make_pair(-1, -1);
    }
  }
  if (n > 1) {
    twin *= n;
    if (twin > 100000) {
      return make_pair(-1, -1);
    }
    betterN *= n;
  }
  return make_pair(twin, betterN);
}
int main(int argc, const char* argv[]) {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long a, b, c, d;
  cin >> a >> b >> c >> d;
  if (a - 1 == b && c == 0 && d == 0) {
    cout << "YES" << endl;
    for (long long i = 0; i <= b - 1; i++) {
      cout << "0 1 ";
    }
    cout << "0" << endl;
    return 0;
  }
  if (a == 0 && b == 0 && c + 1 == d) {
    cout << "YES" << endl;
    for (long long i = 0; i <= c - 1; i++) {
      cout << "3 2 ";
    }
    cout << "3" << endl;
    return 0;
  }
  if (abs(a + c - b - d) >= 2) {
    cout << "NO" << endl;
    return 0;
  }
  vector<long long> solution;
  long long left = a;
  long long right = d;
  long long mid1 = b - left;
  long long mid2 = c - right;
  if (left < 0 || right < 0 || (abs(mid1, mid2) > 1) || mid1 < 0 || mid2 < 0) {
    cout << "NO" << endl;
    return 0;
  }
  if (mid1 < mid2) {
    while (left) {
      solution.push_back(0);
      solution.push_back(1);
      left--;
    }
    solution.push_back(2);
    while (right) {
      solution.push_back(3);
      solution.push_back(2);
      right--;
    }
    while (mid1) {
      solution.push_back(1);
      solution.push_back(2);
      mid1--;
    }
  } else if (mid1 == mid2) {
    while (left) {
      solution.push_back(0);
      solution.push_back(1);
      left--;
    }
    while (right) {
      solution.push_back(2);
      solution.push_back(3);
      right--;
    }
    while (mid1) {
      solution.push_back(2);
      solution.push_back(1);
      mid1--;
    }
  } else {
    while (right) {
      solution.push_back(3);
      solution.push_back(2);
      right--;
    }
    solution.push_back(1);
    while (left) {
      solution.push_back(0);
      solution.push_back(1);
      left--;
    }
    while (mid2) {
      solution.push_back(2);
      solution.push_back(1);
      mid2--;
    }
  }
  cout << "YES" << endl;
  for (long long j = 0; j <= solution.size() - 1; j++) {
    if (j != 0) {
      cout << " ";
    }
    cout << solution[j];
  }
  cout << endl;
  return 0;
}