<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 primes[10001],mod=1e9+7,maxn=10001;
    void sieve(){
        primes[1]=1;
        for(int i=2;i<maxn;i++)primes[i]=i;
        for(int i=2;i*i<maxn;i++){
            if(primes[i]==i){
                for(int j=i*i;j<maxn;j+=i){
                    if(primes[j]==j)primes[j]=i;
                }
            }
        }
    }

    vector<int> factors(int x){
        vector<int>res;
        unordered_map<int,int> cnt;
        while(x!=1){
            cnt[primes[x]]++;
            x=x/primes[x];
        }
        for(auto it : cnt)res.push_back(it.second);
        return res;
    }

    vector<int> waysToFillArray(vector<vector<int>>& queries) {
        vector<int>ans;
        sieve();

        long c[10040][34]={0};
        for(int i=0;i<10040;i++)c[i][0]=1;
        for(int n=1;n<10040;n++){
            for(int k=1;k<34;k++){
                c[n][k]= (c[n-1][k-1] + c[n-1][k])%mod;
            }
        }

        for(auto &it: queries){
            int n=it[0], k=it[1];
            auto cnt=factors(k);
            long curr=1;

            for(int f :cnt){
                curr=(curr*c[n+f-1][f])%mod;
                
            }
           
            ans.push_back(curr);
        }
        return ans;
    }
};
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