<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>

#define md 1000000007
typedef long long ll;
vector<ll> f;
vector<map<ll, ll>> sv;


class Solution {
public:

    void fact()
    {
        f.resize(10010);
        f[0] = 1;
        for (int i = 1; i < 10010; i++) f[i] = (f[i-1] * i) % md;
    }

    void sieve()
    {
        ll n = 10010;
        sv.resize(n + 10);
        for (ll i = 2; i < n; i++)
        {
            if (sv[i].size() == 0) for (ll j = 1, x = i; x < n; j++, x *= i) sv[x][i] = j;
            if (sv[i].size() == 1) 
            {
                ll p = sv[i].begin()->first, tmp = sv[i].begin()->second;
                for (ll j = 1; i * j <= n; j++)
                {
                    if (!(j % p)) continue;
                    sv[i*j][p] = tmp;
                }
            }
        }
    }



    ll pwr(ll a, ll b)
    {
        if (!b) return 1;
        ll t = pwr(a, b >> 1);
        t *= t;
        t %= md;
        if (b & 1) t *= a;
        t %= md;
        return t;
    }

    ll inv(ll n)
    {
        return pwr(n, md - 2);
    }

    ll C(ll n, ll k)
    {
        return (((f[n] * inv(f[k])) % md) * inv(f[n-k])) % md;
    }

    
    int prodArr(int n, int k)
    {
        map<ll, ll>::iterator it = sv[k].begin();
        ll ans = 1;

        while (it != sv[k].end())
        {
            ans *= C(n + (it->second) - 1, n - 1);
            ans %= md;
            it++;
        }

        return ans;
    }


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

        for (ll i = 0; i < queries.size(); i++) ans.push_back(prodArr(queries[i][0], queries[i][1]));

        return ans;
    }
};
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