<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:

    long long mod = 1e9 + 7;
    vector<long long> factorial;

    long long modExp( long long a, long long b ){
        if( b == 0 ) return 1;
        a = a % mod;
        if( b & 1 ) return (a * modExp( a, b - 1 )) % mod;
        long long temp = modExp( a, b/2 ) % mod;
        return ( temp * temp ) % mod;
    }

    long long inverseModulo( int a ){
        return modExp( a, mod - 2 ) % mod;
    }

    void init( int n ){
        factorial.resize(n);
        factorial[0] = 1;
        for( int i = 1; i < n; i++ ){
            factorial[i] = ( i * factorial[i-1] ) % mod;
        }
    }

    long long ncr( int n, int r ){
        // cout << n << " " << r << " " << n - r << endl;
        long long num = factorial[n] % mod;
        long long den1 = factorial[r];
        long long den2 = factorial[n-r];
        // cout << num << " " << den1 << " " << den2 << endl << endl;
        // return 1;

        return ((num * inverseModulo(den1)) % mod * inverseModulo(den2)) % mod;
    }

    long long solve( int len, int prod ){
        vector<int> pf;
        int count2 = 0;
        long long ans = 1;
        int tempProd = prod;
        while( prod % 2 == 0 ){
            count2++;
            prod /= 2;
        }
        ans = ( ans * ncr( count2 + len - 1, count2 ) ) % mod;
        for( int i = 3; i * i <= tempProd; i += 2 ){
            if( tempProd % i == 0 ){
                int count = 0;
                while( prod % i == 0 ){
                    count++;
                    prod /= i;
                }
                ans = ( ans * ncr( count + len - 1, count ) ) % mod;
            }
        }
        if( prod != 1 ) {
            ans = ( ans * ncr( len, 1 ) ) % mod;
        }
        return ans;
    }

    vector<int> waysToFillArray(vector<vector<int>>& queries) {
        
        init( 1e5 + 5 );

        vector<int> ansv;
        // return ansv;

        for( auto q : queries ){
            int len = q[0], prod = q[1];
            long long ans = solve( len, prod );
            ansv.push_back(ans);
        }

        return ansv;

    }
};
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