<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>

constexpr int MOD = 1e9 + 7;

class Solution {
public:
    vector<int> waysToFillArray(vector<vector<int>>& queries) {
        int maxN = queries[0][0];
        int maxK = queries[0][1];

        for (int i = 1; i < queries.size(); i++) {
            maxN = max(maxN, queries[i][0]);
            maxK = max(maxK, queries[i][1]);
        }

        vector<int> divisors(toDivisorIndex(maxK) + 1); // divisors[i] = p -> p is a divisor of x = 2 * i + 3  (i = (x - 3) / 2)

        for (int i = 0; i < divisors.size(); i++) {
            divisors[i] = fromDivisorIndex(i);
        }

        int maxP = floor(sqrt(maxK)) + 1;
        for (int i = 0, p = 3; p < maxP; i++, p += 2) {
            if (divisors[i] == p) {
                for (int j = i + p; j < divisors.size(); j += p) {
                    divisors[j] = p;
                }
            }
        }


        int maxDivisors = floor(log2(maxK));
        vector<vector<int>> dp(maxN + 1, vector<int>(maxDivisors + 1, 1)); // dp[n][d] = ways to arrange d prime factors among n values
        // dp[n][d] = dp[n - 1][d] + dp[n - 1][d - 1] + ... + dp[n - 1][0]
        //          = dp[n - 1][d] + dp[n][d - 1]
        // dp[n = 1][d] = dp[n][d = 0] = 1

        for (int n = 2; n < dp.size(); n++) {
            for (int d = 1; d < dp[n].size(); d++) {
                dp[n][d] = (dp[n - 1][d] + dp[n][d - 1]) % MOD;
            }
        }


        vector<int> results(queries.size());

        for (int i = 0; i < queries.size(); i++) {
            int n = queries[i][0];
            int k = queries[i][1];
            long long ways = 1;

            while (k > 1) {
                int divisor = (k % 2 == 0 ? 2 : divisors[toDivisorIndex(k)]);
                int currentDivisorCount = 0;

                do {
                    k /= divisor;
                    currentDivisorCount++;
                } while (k % divisor == 0);

                ways = (ways * dp[n][currentDivisorCount]) % MOD;
            }

            results[i] = ways;
        }

        return results;
    }

private:
    int toDivisorIndex(int x) {
        return (x - 3) / 2;
    }

    int fromDivisorIndex(int i) {
        return 2 * i + 3;
    }
};

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