codeforces Round #641 div.2 A. Orac and Factorstime

                     **A. Orac and Factorstime**
                        limit per test2 second
              smemory limit per test256 megabytes
                     input standard input
                     output standard output

Orac is studying number theory, and he is interested in the properties of divisors.For two positive integers aa and bb, aa is a divisor of bb if and only if there exists an integer cc, such that a⋅c=ba⋅c=b.For n≥2n≥2, we will denote as f(n)f(n) the smallest positive divisor of nn, except 11.For example, f(7)=7,f(10)=2,f(35)=5f(7)=7,f(10)=2,f(35)=5.For the fixed integer nn, Orac decided to add f(n)f(n) to nn.For example, if he had an integer n=5n=5, the new value of nn will be equal to 1010. And if he had an integer n=6n=6, nn will be changed to 88.Orac loved it so much, so he decided to repeat this operation several times.Now, for two positive integers nn and kk, Orac asked you to add f(n)f(n) to nn exactly kk times (note that nn will change after each operation, so f(n)f(n) may change too) and tell him the final value of nn.For example, if Orac gives you n=5n=5 and k=2k=2, at first you should add f(5)=5f(5)=5 to n=5n=5, so your new value of nn will be equal to n=10n=10, after that, you should add f(10)=2f(10)=2 to 1010, so your new (and the final!) value of nn will be equal to 1212.Orac may ask you these queries many times.InputThe first line of the input is a single integer t (1≤t≤100)t (1≤t≤100): the number of times that Orac will ask you.Each of the next tt lines contains two positive integers n,k (2≤n≤106,1≤k≤109)n,k (2≤n≤106,1≤k≤109), corresponding to a query by Orac.It is guaranteed that the total sum of nn is at most 106106.OutputPrint tt lines, the ii-th of them should contain the final value of nn in the ii-th query by Orac.
Example
input
5 1
8 2
3 4
output
12
12
NoteIn the first query, n=5n=5 and k=1k=1. The divisors of 55 are 11 and 55, the smallest one except 11 is 55. Therefore, the only operation adds f(5)=5f(5)=5 to 55, and the result is 1010.In the second query, n=8n=8 and k=2k=2. The divisors of 88 are 1,2,4,81,2,4,8, where the smallest one except 11 is 22, then after one operation 88turns into 8+(f(8)=2)=108+(f(8)=2)=10. The divisors of 1010 are 1,2,5,101,2,5,10, where the smallest one except 11 is 22, therefore the answer is 10+(f(10)=2)=1210+(f(10)=2)=12.In the third query, nn is changed as follows: 3→6→8→10→123→6→8→10→12.

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
int main()
{
 int t;
 long long int n,k;
 cin>>t;
 while(t--){
  cin>>n>>k;
   if(n%2==0) n=n+2*k;
   else if(n%2==1){
    for(int j=2;j<=n;j++)
     if(n%j==0) {n=n+j+2*k-2;break;} 
   } 
   cout<<n<<endl; 
  }
 return 0;
 }