<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>

const int N=2e4+1;
class Solution {
public:
    long mod=1e9+7;
    long fac[N],fac_inv[N];

    long pw(long a, long b){
        if(b==0){
            return 1;
        }
        long c=pw(a,b/2);
        return ((c*c)%mod *(b%2 ? a :1))%mod;
    }

    long Cnr(long n, long r){
        if(r>n || r<0){
            return 0;
        }
        return ((fac[n]*fac_inv[r])%mod * fac_inv[n-r])%mod;
    }


    vector<int> waysToFillArray(vector<vector<int>>& queries) {
        
        fac[0]=1;
        for(int i=1;i<N;++i){
            fac[i]=(fac[i-1]*i)%mod;
        }
        fac_inv[N-1]=pw(fac[N-1],mod-2);
        for(int i=N-2;i>=0;--i){
            fac_inv[i]=((i+1)*fac_inv[i+1])%mod;
        }
        vector<int>ans;
        for(auto&it:queries){
            long res=1;
            for(int i=2;i*i<=it[1];++i){
                if(it[1]%i==0){
                    int cnt=0;
                    while(it[1]%i==0){
                        it[1]/=i;
                        cnt++;
                    }
                    res=(res*Cnr(cnt+it[0]-1,it[0]-1))%mod;
                }
            }
            if(it[1]>1){
                res=(res*Cnr(1+it[0]-1,it[0]-1))%mod;
            }
            ans.push_back(res);
        }
        return ans;
    }
};
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