<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>

int N = 10020;
int MOD = 1e9 + 7;
long f[10020];
long g[10020];
vector<int> p[10020];

long qmi(long a, long k, long p) {
    long res = 1;
    while (k != 0) {
        if ((k & 1) == 1) {
            res = res * a % p;
        }
        k >>= 1;
        a = a * a % p;
    }
    return res;
}

int init = []() {
    f[0] = 1;
    g[0] = 1;
    for (int i = 1; i < N; ++i) {
        f[i] = f[i - 1] * i % MOD;
        g[i] = qmi(f[i], MOD - 2, MOD);
        int x = i;
        for (int j = 2; j <= x / j; ++j) {
            if (x % j == 0) {
                int cnt = 0;
                while (x % j == 0) {
                    ++cnt;
                    x /= j;
                }
                p[i].push_back(cnt);
            }
        }
        if (x > 1) {
            p[i].push_back(1);
        }
    }
    return 0;
}();

int comb(int n, int k) {
    return (f[n] * g[k] % MOD) * g[n - k] % MOD;
}

class Solution {
public:
    vector<int> waysToFillArray(vector<vector<int>>& queries) {
        vector<int> ans;
        for (auto& q : queries) {
            int n = q[0], k = q[1];
            long long t = 1;
            for (int x : p[k]) {
                t = t * comb(x + n - 1, n - 1) % MOD;
            }
            ans.push_back(t);
        }
        return ans;
    }
};
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