1. Abridged Problem Statement
Given a positive integer n (1 ≤ n ≤ 10^12), determine the number of distinct finite abelian groups (up to isomorphism) of order n. Output this count.

2. Detailed Editorial

**Theory**
- **Classification**. By the Fundamental Theorem of Finite Abelian Groups, any finite abelian group of order n factors uniquely (up to isomorphism) into a direct sum of p-power cyclic groups, one for each prime dividing n. Concretely, if
  n = ∏ p_i^{e_i},
  then every abelian group of order n is a direct sum over i of abelian p_i-groups of order p_i^{e_i}.
- **Counting p-groups**. The number of abelian p-groups of order p^e is equal to the number of integer partitions of e. Denote by p(e) the partition function: the count of ways to write e as a sum of positive integers in nonincreasing order.
- **Multiplicativity**. Different primes give independent choices, so the total number is
  Answer = ∏_{i} p(e_i).

**Implementation Steps**
1. **Prime factorization** of n (up to 10^12). A simple trial division up to √n (∼10^6) suffices. Record each exponent e_i.
2. **Partition numbers** up to max(e_i). Since max(e_i) ≤ log₂(10^12) < 40, we can safely precompute partitions up to 100. We use Euler's recurrence with generalized pentagonal numbers:
   p(0)=1; for m>0,
   p(m) = ∑_{k=1..∞} (−1)^{k−1} [p(m − k(3k−1)/2) + p(m − k(3k+1)/2)],
   stopping when arguments go negative.
3. **Multiply** the relevant p(e_i) together.
4. **Output** the product.

**Complexities**
- Factorization: O(√n).
- Partition precompute (up to N): O(N √N).
- Overall fits easily in the time limit for n ≤ 10^12.

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;
};

int64_t n;

void read() { cin >> n; }

vector<int64_t> compute_partitions(int max_n) {
    vector<int64_t> p(max_n + 1, 0);
    p[0] = 1;

    for(int n = 1; n <= max_n; ++n) {
        int64_t sum = 0;
        int k = 1;
        while(true) {
            int pent1 = (k * (3 * k - 1)) / 2;
            int pent2 = ((-k) * (3 * (-k) - 1)) / 2;
            if(pent1 > n && pent2 > n) {
                break;
            }

            int sign = (k % 2 == 1) ? 1 : -1;
            if(pent1 <= n) {
                sum += sign * p[n - pent1];
            }

            if(pent2 <= n && pent2 != pent1) {
                sum += sign * p[n - pent2];
            }

            ++k;
        }

        p[n] = sum;
    }

    return p;
}

void solve() {
    // By the fundamental theorem of finite abelian groups and the Chinese
    // remainder theorem, the number of abelian groups of order n is
    // multiplicative over the prime factorization: for a prime power p^e the
    // count of distinct decompositions equals the number of integer partitions
    // of e. So factorize n, and multiply together the partition counts of each
    // exponent. compute_partitions uses Euler's pentagonal number recurrence to
    // build the partition function up to 100 (the max exponent for n <= 10^12
    // is 39, since 2^40 > 10^12). The leftover factor after trial division up
    // to sqrt(n) is a single prime with exponent 1, contributing p(1) = 1.

    vector<int64_t> partitions = compute_partitions(100);
    int64_t result = 1;
    for(int64_t p = 2; p * p <= n; ++p) {
        if(n % p == 0) {
            int e = 0;
            while(n % p == 0) {
                e++;
                n /= p;
            }

            result *= partitions[e];
        }
    }

    if(n > 1) {
        result *= partitions[1];
    }

    cout << result << '\n';
}

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

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

    return 0;
}
```

4. Python Solution

```python
import sys
import math

def compute_partitions(max_n):
    """
    Compute partition numbers p(0..max_n) using Euler's
    recurrence with generalized pentagonal numbers.
    """
    p = [0] * (max_n + 1)
    p[0] = 1  # Base case

    for m in range(1, max_n + 1):
        total = 0
        k = 1
        while True:
            # Generalized pentagonal numbers:
            pent1 = k * (3*k - 1) // 2
            pent2 = k * (3*k + 1) // 2
            if pent1 > m and pent2 > m:
                break
            sign = 1 if (k % 2 == 1) else -1

            if pent1 <= m:
                total += sign * p[m - pent1]
            if pent2 <= m:
                total += sign * p[m - pent2]

            k += 1
        p[m] = total

    return p

def main():
    # Read n from stdin
    n = int(sys.stdin.readline())
    temp = n
    result = 1

    # Factor n by trial division
    exponents = []
    # Check divisor 2 separately for slight speedup
    if temp % 2 == 0:
        cnt = 0
        while temp % 2 == 0:
            temp //= 2
            cnt += 1
        exponents.append(cnt)

    # Check odd divisors up to sqrt(temp)
    f = 3
    while f * f <= temp:
        if temp % f == 0:
            cnt = 0
            while temp % f == 0:
                temp //= f
                cnt += 1
            exponents.append(cnt)
        f += 2

    # If remainder > 1, it's a prime factor with exponent 1
    if temp > 1:
        exponents.append(1)

    # Precompute partitions up to the largest exponent
    max_e = max(exponents, default=0)
    partitions = compute_partitions(max_e)

    # Multiply partition(exponent) for each prime's exponent
    for e in exponents:
        result *= partitions[e]

    # Print the result
    print(result)

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

5. Compressed Editorial
- Factor n = ∏ pᵢ^{eᵢ}.
- By classification, # of abelian p-groups of order pᵉ = partition(e).
- Overall answer = ∏ partition(eᵢ).
- Compute partition numbers up to max e via Euler's pentagonal recurrence in O(N√N).
- Factor n by trial division in O(√n).
- Multiply and output.
