<p>You are given a 2D integer array, <code>queries</code>. For each <code>queries[i]</code>, where <code>queries[i] = [n<sub>i</sub>, k<sub>i</sub>]</code>, find the number of different ways you can place positive integers into an array of size <code>n<sub>i</sub></code> such that the product of the integers is <code>k<sub>i</sub></code>. As the number of ways may be too large, the answer to the <code>i<sup>th</sup></code> query is the number of ways <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>

<p>Return <em>an integer array </em><code>answer</code><em> where </em><code>answer.length == queries.length</code><em>, and </em><code>answer[i]</code><em> is the answer to the </em><code>i<sup>th</sup></code><em> query.</em></p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> queries = [[2,6],[5,1],[73,660]]
<strong>Output:</strong> [4,1,50734910]
<strong>Explanation:</strong>&nbsp;Each query is independent.
[2,6]: There are 4 ways to fill an array of size 2 that multiply to 6: [1,6], [2,3], [3,2], [6,1].
[5,1]: There is 1 way to fill an array of size 5 that multiply to 1: [1,1,1,1,1].
[73,660]: There are 1050734917 ways to fill an array of size 73 that multiply to 660. 1050734917 modulo 10<sup>9</sup> + 7 = 50734910.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> queries = [[1,1],[2,2],[3,3],[4,4],[5,5]]
<strong>Output:</strong> [1,2,3,10,5]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= queries.length &lt;= 10<sup>4</sup> </code></li>
	<li><code>1 &lt;= n<sub>i</sub>, k<sub>i</sub> &lt;= 10<sup>4</sup></code></li>
</ul>

#include <bits/stdc++.h>
using namespace std;

class Solution {
public:
    typedef long long ll;
    const int MOD = 1e9 + 7;
    static const int MAX = 20010; // Maximum possible value for n + a_i -1

    // Precomputed factorial and inverse factorial
    vector<long long> fact;
    vector<long long> inv_fact;
    vector<int> spf; // Smallest Prime Factor

    // Function to compute a^b mod MOD
    ll power_mod(ll a, ll b){
        ll res = 1;
        a %= MOD;
        while(b > 0){
            if(b & 1){
                res = res * a % MOD;
            }
            a = a * a % MOD;
            b >>= 1;
        }
        return res;
    }

    // Function to compute combination C(n, k) mod MOD
    ll comb(int n, int k){
        if(n < k || k < 0) return 0;
        return fact[n] * inv_fact[k] % MOD * inv_fact[n - k] % MOD;
    }

    // Function to factorize a number using SPF sieve
    vector<pair<int, int>> factorize(int x){
        vector<pair<int, int>> factors;
        while(x > 1){
            int p = spf[x];
            int cnt = 0;
            while(x % p == 0){
                cnt++;
                x /= p;
            }
            factors.emplace_back(p, cnt);
        }
        return factors;
    }

    // Constructor to initialize precomputations
    Solution(){
        // Initialize factorial
        fact.assign(MAX, 1);
        for(int i = 1; i < MAX; ++i){
            fact[i] = fact[i - 1] * i % MOD;
        }

        // Initialize inverse factorial
        inv_fact.assign(MAX, 1);
        inv_fact[MAX - 1] = power_mod(fact[MAX - 1], MOD - 2); // Fermat's Little Theorem
        for(int i = MAX - 2; i >= 0; --i){
            inv_fact[i] = inv_fact[i + 1] * (i + 1) % MOD;
        }

        // Initialize SPF (Smallest Prime Factor) sieve
        spf.assign(MAX, 0);
        for(int i = 2; i < MAX; ++i){
            if(spf[i] == 0){
                spf[i] = i;
                for(int j = 2 * i; j < MAX; j += i){
                    if(spf[j] == 0){
                        spf[j] = i;
                    }
                }
            }
        }
    }

    vector<int> waysToFillArray(vector<vector<int>>& queries) {
        vector<int> ans;
        for(auto &q : queries){
            int n = q[0];
            int k = q[1];
            if(k == 1){
                ans.push_back(1);
                continue;
            }
            // Factorize k
            vector<pair<int, int>> factors = factorize(k);
            // Compute the number of ways
            ll res = 1;
            for(auto &[p, cnt] : factors){
                res = res * comb(n + cnt - 1, cnt) % MOD;
            }
            ans.push_back(res);
        }
        return ans;
    }
};

68000