1. Abridged Problem Statement
Given a positive integer n (≤10000), express n as a sum of super-prime numbers using as few terms as possible.
A super-prime is a prime whose position in the sequence of all primes is itself prime (e.g., 3 is the 2nd prime and 2 is prime, so 3 is a super-prime; 7 is the 4th prime and 4 is not prime, so 7 is not).
If no such representation exists, output 0. Otherwise, output:
• First line: the minimal count I of super-primes in the sum.
• Second line: I super-primes in non-increasing order that sum to n.

2. Detailed Editorial
Overview
This is a classic "coin-change" minimization problem where the "coins" are all super-prime numbers ≤n. We need the minimum number of coins whose sum is exactly n, and then recover one optimal combination sorted non-increasingly.

Step A: Generate super-primes
1. Sieve all primes (the reference precomputes up to 10^6, a safe upper bound).
2. Maintain a counter cnt of primes seen so far; when you encounter the cnt-th prime i, if cnt is itself prime (checked with the same sieve, since cnt ≤ i), mark i as a super-prime.

Step B: Compute minimum-coins DP
Define dp[x] = minimum number of super-primes summing to x (or –1 if unreachable).
Initialize dp[0] = 0, dp[1..n] = –1.
For each super-prime s and for each sum j from s..n:
  if dp[j–s] != –1, then dp[j] = min(dp[j], dp[j–s] + 1) (initializing if unset).

Step C: Check feasibility
• If dp[n] == –1, print 0.
• Otherwise, reconstruct one solution by walking backwards:
  While current sum > 0, find any super-prime s such that dp[sum–s] == dp[sum]–1; append s, decrease sum by s.

Step D: Output
Print dp[n], then sort the collected coins in descending order and print them.

Complexities
• Sieve: O(M log log M).
• DP: O(n × #superprimes). For n=10000 this is comfortably fast.

3. C++ Solution

```cpp
#include <bits/stdc++.h>

using namespace std;

template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second;
}

template<typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
    return in >> x.first >> x.second;
}

template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    for(auto& x: a) {
        in >> x;
    }
    return in;
};

template<typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
    for(auto x: a) {
        out << x << ' ';
    }
    return out;
};

vector<int> super_primes(int n) {
    vector<bool> is_prime(n + 1, true);
    is_prime[0] = is_prime[1] = false;

    int cnt = 0;
    vector<int> res;
    for(int i = 2; i <= n; i++) {
        if(is_prime[i]) {
            cnt++;
            if(is_prime[cnt]) {
                res.push_back(i);
            }

            for(int64_t j = i * 1ll * i; j <= n; j += i) {
                is_prime[j] = false;
            }
        }
    }

    return res;
}

vector<int> sp;
int n;

void read() { cin >> n; }

void solve() {
    // A super-prime is the k-th prime where k itself is prime. We precompute
    // all super-primes by sieving and checking that the prime's index is
    // prime. Then this is an unbounded coin-change: dp[j] = fewest super-primes
    // summing to j (-1 if impossible). We reconstruct one optimal multiset by
    // greedily peeling off a super-prime that keeps dp on an optimal path, and
    // print the summands in non-increasing order.

    vector<int> dp(n + 1, -1);
    dp[0] = 0;
    for(int i = 0; i < (int)sp.size(); i++) {
        for(int j = sp[i]; j <= n; j++) {
            if(dp[j - sp[i]] != -1) {
                if(dp[j] == -1 || dp[j - sp[i]] + 1 < dp[j]) {
                    dp[j] = dp[j - sp[i]] + 1;
                }
            }
        }
    }

    if(dp[n] == -1) {
        cout << 0 << '\n';
        return;
    }

    vector<int> ans;
    while(n > 0) {
        for(int i = 0; i < (int)sp.size(); i++) {
            if(n >= sp[i] && dp[n - sp[i]] == dp[n] - 1) {
                ans.push_back(sp[i]);
                n -= sp[i];
                break;
            }
        }
    }

    cout << ans.size() << '\n';
    sort(ans.rbegin(), ans.rend());
    cout << ans << '\n';
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    sp = super_primes((int)1e6);

    int T = 1;
    // cin >> T;
    for(int test = 1; test <= T; test++) {
        read();
        // cout << "Case #" << test << ": ";
        solve();
    }

    return 0;
}
```

4. Python Solution with Detailed Comments
```python
import sys
def main():
    data = sys.stdin.read().strip().split()
    if not data:
        return
    n = int(data[0])

    # Sieve primes up to n
    is_prime = [True] * (n + 1)
    is_prime[0] = is_prime[1] = False
    primes = []
    for i in range(2, n + 1):
        if is_prime[i]:
            primes.append(i)
            for j in range(i*i, n+1, i):
                is_prime[j] = False

    # Identify super-primes: the k-th prime is super if k is prime
    sp = []
    maxk = len(primes)
    is_prime_k = [True] * (maxk + 1)
    if maxk >= 0:
        is_prime_k[0] = is_prime_k[1] = False
    for i in range(2, maxk + 1):
        if is_prime_k[i]:
            for j in range(i*i, maxk+1, i):
                is_prime_k[j] = False

    for idx, p in enumerate(primes, start=1):
        if idx <= maxk and is_prime_k[idx]:
            sp.append(p)

    # dp[x] = min number of super-primes summing to x, or inf if impossible
    INF = 10**9
    dp = [INF] * (n + 1)
    dp[0] = 0

    # Unbounded coin change over sp[]
    for coin in sp:
        for s in range(coin, n+1):
            if dp[s-coin] + 1 < dp[s]:
                dp[s] = dp[s-coin] + 1

    if dp[n] >= INF:
        print(0)
        return

    # Reconstruct solution
    res = []
    cur = n
    while cur > 0:
        for coin in sp:
            if cur >= coin and dp[cur-coin] == dp[cur] - 1:
                res.append(coin)
                cur -= coin
                break

    # Output result
    print(len(res))
    res.sort(reverse=True)
    print(*res)

if __name__ == "__main__":
    main()
```

5. Compressed Editorial
1. Sieve primes; collect those whose index in the prime list is prime → super-primes.
2. Use unbounded-knapsack DP: dp[0]=0, dp[x]=min(dp[x−s]+1) over super-primes s.
3. If dp[n] is unreachable, print 0. Otherwise backtrack one optimal solution and print count and coins sorted descending.
