<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 {
    // let's change question 
    // find number of ways to distribute k elements in size n array 
    // k is atmax 20 
    // such that everyone have >= 0 elements and none should be left unused 
    int MOD = 1e9 + 7; 
    int dp[10004][21]; 
    int recursive(int n,int k){ 
        if(k == 0) return 1; // this is a valid way of distribution
        if(n == 0) return 0; 
        if(dp[n][k] != -1) return dp[n][k]; 
        return dp[n][k] = (recursive(n,k-1) + recursive(n-1,k))%MOD;
    }
public:
    vector<int> waysToFillArray(vector<vector<int>>& queries) {
        for(int i=0;i<10004;i+=1) for(int j=0;j<21;j+=1) dp[i][j] = -1; recursive(10003,20); 
        vector<int> res; 
        for(auto &q: queries){
            int n = q[0]; 
            int k = q[1]; 
            // find prime factorizatio of k then work forward 
            map<int,int> mp; 
            for(int i=2;i*i<=k;i+=1){
                while(q[1]%i == 0) q[1]/=i,mp[i] += 1; 
            }
            if(q[1] != 1) mp[q[1]] += 1; 
            long long ans = 1; 
            for(auto &[elm,freq]: mp){
                ans *= dp[n][freq]; 
                ans %= MOD; 
            }
            res.push_back(ans); 
        }
        return res;
    }
};
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