<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>

class Solution {
public:
    int small_prime[10001],m=1e9+7,MAX=10001;
    void sieve() {
        small_prime[1]=1;
        for(int i=2; i<MAX; i++) small_prime[i]=i;
        for(int i=4; i<MAX; i+=2) small_prime[i]=2;
        for(int i=3; i*i<MAX; i++) {
            if(small_prime[i]==i) {
                for(int j=i*i; j<10001; j+=i)
                    if(small_prime[j]==j) small_prime[j]=i;
            }
        }
    }
    vector<int> factors(int x) {
        vector<int> arr;
        unordered_map<int,int> freq;
        while(x!=1) {
            freq[small_prime[x]]++;
            x=x/small_prime[x];
        }
        for(auto it:freq) arr.push_back(it.second);
        return arr;
    }
    vector<int> waysToFillArray(vector<vector<int>>& queries) {
        vector<int> ans;
        sieve();
        long comb[10050][33]={0};
        for(int j=0; j!=10050; j++) comb[j][0]=1;
        for(int i=1; i!=10050; i++) {
            for(int j=1; j!=33; j++)
                comb[i][j]=(comb[i-1][j-1]+comb[i-1][j])%m;
        }
        for(auto& it:queries) {
            int n=it[0],k=it[1];
            auto freq=factors(k);
            long curr=1;
            for(int f:freq)
                curr=(curr*comb[n+f-1][f])%m;
            ans.push_back(curr);
        }
        return ans;
    }
};
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