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 up to n  
1. Sieve all primes up to n.  
2. Record each prime in order; when you encounter the k-th prime, if k is prime, mark that prime 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(n log log n)  
• Number of primes ≤n is about n/log n; number of super-primes is smaller.  
• DP: O(n × #superprimes). For n=10000 this is comfortably fast.

3. Provided C++ Solution with Detailed Comments  
```cpp
#include <bits/stdc++.h>
using namespace std;

// Overloaded output for a pair<int,int>
template<typename T1, typename T2>
ostream &operator<<(ostream &out, const pair<T1, T2> &x) {
    // Print as "first second"
    return out << x.first << ' ' << x.second;
}

// Overloaded input for a pair<int,int>
template<typename T1, typename T2>
istream &operator>>(istream &in, pair<T1, T2> &x) {
    return in >> x.first >> x.second;
}

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

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

// Generate all super-primes up to n.
// A super-prime is a prime whose index in the prime list is itself prime.
vector<int> super_primes(int n) {
    vector<bool> is_prime(n + 1, true);
    is_prime[0] = is_prime[1] = false;
    int cnt = 0;  // Count of primes found so far
    vector<int> super_primes;

    for(int i = 2; i <= n; i++) {
        if(is_prime[i]) {
            cnt++;  // i is the cnt-th prime
            // If cnt is prime, i is a super-prime
            if(is_prime[cnt]) {
                super_primes.push_back(i);
            }
            // Mark multiples of i as non-prime
            for(int64_t j = 1LL * i * i; j <= n; j += i) {
                is_prime[j] = false;
            }
        }
    }
    return super_primes;
}

vector<int> sp;  // List of all super-primes up to limit
int n;          // Target sum

// Read input
void read() {
    cin >> n;
}

// Solve one test
void solve() {
    // dp[x] = minimum number of super-primes summing to x, or -1 if impossible
    vector<int> dp(n + 1, -1);
    dp[0] = 0;  // Zero coins needed to make sum 0

    // Unbounded knapsack: try each super-prime as a coin
    for(int s: sp) {
        for(int j = s; j <= n; j++) {
            if(dp[j - s] != -1) {
                int candidate = dp[j - s] + 1;
                if(dp[j] == -1 || candidate < dp[j]) {
                    dp[j] = candidate;
                }
            }
        }
    }

    // If dp[n] is still -1, no representation
    if(dp[n] == -1) {
        cout << 0 << '\n';
        return;
    }

    // Reconstruct one optimal solution
    vector<int> ans;
    int cur = n;
    while(cur > 0) {
        // Find a super-prime s that leads to dp[cur]
        for(int s: sp) {
            if(cur >= s && dp[cur - s] == dp[cur] - 1) {
                ans.push_back(s);
                cur -= s;
                break;
            }
        }
    }

    // Output the count and the sorted list in descending order
    cout << ans.size() << '\n';
    sort(ans.rbegin(), ans.rend());
    cout << ans << '\n';
}

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

    // Precompute super-primes up to 1e6 (safe upper bound)
    sp = super_primes(1000000);

    read();
    solve();
    return 0;
}
```

4. Python Solution with Detailed Comments  
```python
import sys
import threading
def main():
    sys.setrecursionlimit(10**7)
    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 = []
    # Recompute primality up to len(primes)
    maxk = len(primes)
    is_prime_k = [True] * (maxk + 1)
    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 up to n; 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 infinite, print 0. Otherwise backtrack one optimal solution and print count and coins sorted descending.