<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>

long long ncr[10014][14] = {1}, mod = 1e9 + 7;
int primes[] = {2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97};
class Solution {
public:
    vector<int> waysToFillArray(vector<vector<int>>& queries) {
        if(ncr[1][1] == 0) {
            for(int n=1; n<10014; n++) {
                for(int r=0; r<14; r++) {
                    ncr[n][r] = (r==0) ? 1 : (ncr[n-1][r-1] + ncr[n-1][r]) % mod;
                }
            }
        }
        vector<int> ans(queries.size(),1);
        for(int i=0; i<queries.size(); i++) {
            int n = queries[i][0], pro = queries[i][1];
            for(auto prime: primes) {
                int r = 0;
                while(pro % prime == 0) {
                    r++;
                    pro /= prime;
                }
                ans[i] = (ans[i] * 1LL * ncr[n+r-1][r]) % mod;
            }
            if(pro != 1) ans[i] = (ans[i] * 1LL* n) % mod;
        }
        return ans;
    }
};
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