<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 MOD=1e9+7;
const int N = 10060; //CHANGE ACCORDINGLY
const int mod = 1000000007;
#define ll long long
class Solution {
public:
    ll fact[N];
    ll invFact[N];
    ll fast_pow(ll a, ll p) {
        int var = 1;
        while (p) {
            if (p % 2 == 0) {
                a = (a * a) % mod;
                p /= 2;
            } else {
                var = (var * a) % mod;
                p--;
            }
        }
        return var;
    }
    int C(int n, int k) {
        if (k > n) {
            return 0;
        }
        return fact[n] * invFact[k] % mod * invFact[n - k] % mod;
    }
    void once() {
        fact[0] = invFact[0] = 1;
        for (int i = 1; i < N; i++) {
            fact[i] = (fact[i - 1] * i) % mod;
            invFact[i] = fast_pow(fact[i], mod - 2);
        }
    }
    vector<int> waysToFillArray(vector<vector<int>>& queries) {
        once();
        vector<int> res;
        for(auto q:queries){
            int n=q[0];
            int num=q[1];
            long long ans=1;
            for(int i=2; i<=sqrt(num); i++){
                int r=0;
                while(num%i==0){
                    num/=i;
                    r++;
                }
                // cout<<r+n-1<<" "<<n-1<<endl;
                // cout<<comb[r+n-1][n-1]<<endl;
                ans*=1ll*C(r+n-1,n-1);
                ans%=MOD;
            }
            if(num!=1){
                ans*=n;
                ans%=MOD;
            }   
            res.push_back(ans);
        }
        return res;
    }
};
34000