# HackerEarth Factors problem solution

In this HackerEarth Factors problem solution, Ben had two numbers M and N. He factorized both the numbers, that is, expressed M and N as a product of prime numbers.
M = (P1A1)*(P2A2 )*(P3A3)*… *(PNAN)
N = (Q1B1)*(Q2B2 )*(Q3B3)*… *(QNBN)

Here, * represents multiplication.
P1, P2 ...PN  are distinct prime numbers.
Q1, Q2 ...QN are distinct prime numbers.
Unfortunately, Ben has lost both the numbers M and N but still he has arrays A and B. He wants to reterive the numbers M and N but he will not be able to. Your task is to determine the minimum and the maximum number of factors their product (that is, M * N) could have. Your task is to tell Ben the minimum and the maximum number of factors that the product of M and N could have.
Since the answer can be large, print modulo 1000000007.

## HackerEarth Factors problem solution.

`#include<bits/stdc++.h>using namespace std;#define FIO ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0)#define m 1000000007#define test ll t; cin>>t; while(t--)typedef long long int ll;int main() {    FIO;    test    {       ll n,i,ans1=1,ans2=1;       cin>>n;       ll a[n],b[n];       for(i=0;i<n;i++){           cin>>a[i];       }       for(i=0;i<n;i++){           cin>>b[i];       }       sort(a,a+n);       sort(b,b+n,greater<ll>());       for(i=0;i<n;i++){           ans1*=(a[i]+b[i]+1);           ans1%=m;           ans2*=(a[i]+1);           ans2%=m;           ans2*=(b[i]+1);           ans2%=m;       }       cout<<ans1<<" "<<ans2<<endl;    }    return 0;}`

### Second solution

`#include <bits/stdc++.h>using namespace std;typedef long long ll;const int N = 1e5 + 14, M = 1e9 + 7;int n;int main() {    ios::sync_with_stdio(0), cin.tie(0);    int t;    cin >> t;    while (t--) {        cin >> n;        int a[n], b[n];        int mx = 1;        for (int i = 0; i < n; ++i) {            cin >> a[i];            mx = (ll) mx * (a[i] + 1) % M;        }        for (int i = 0; i < n; ++i) {            cin >> b[i];            mx = (ll) mx * (b[i] + 1) % M;        }        sort(a, a + n);        sort(b, b + n, greater<int>());        int mn = 1;        for (int i = 0; i < n; ++i)            mn = (ll) mn * (a[i] + b[i] + 1) % M;        cout << mn << ' ' << mx << '\n';    }}`