Orac is studying number theory, and he is interested in the properties of divisors.

For two positive integers a and b, a is a divisor of b if and only if there exists an integer c, such that a⋅ c=b.

For n ≥ 2, we will denote as f(n) the smallest positive divisor of n, except 1.

For example, f(7)=7,f(10)=2,f(35)=5.

For the fixed integer n, Orac decided to add f(n) to n. 

For example, if he had an integer n=5, the new value of n will be equal to 10. And if he had an integer n=6, n will be changed to 8.

Orac loved it so much, so he decided to repeat this operation several times.

Now, for two positive integers n and k, Orac asked you to add f(n) to n exactly k times (note that n will change after each operation, so f(n) may change too) and tell him the final value of n.

For example, if Orac gives you n=5 and k=2, at first you should add f(5)=5 to n=5, so your new value of n will be equal to n=10, after that, you should add f(10)=2 to 10, so your new (and the final!) value of n will be equal to 12.

Orac may ask you these queries many times.

Input

The first line of the input is a single integer t\ (1≤ t≤ 100): the number of times that Orac will ask you.

Each of the next t lines contains two positive integers n,k\ (2≤ n≤ 10^6, 1≤ k≤ 10^9), corresponding to a query by Orac.

It is guaranteed that the total sum of n is at most 10^6. 

Output

Print t lines, the i-th of them should contain the final value of n in the i-th query by Orac.

Example

Input


3
5 1
8 2
3 4


Output


10
12
12

Note

In the first query, n=5 and k=1. The divisors of 5 are 1 and 5, the smallest one except 1 is 5. Therefore, the only operation adds f(5)=5 to 5, and the result is 10.

In the second query, n=8 and k=2. The divisors of 8 are 1,2,4,8, where the smallest one except 1 is 2, then after one operation 8 turns into 8+(f(8)=2)=10. The divisors of 10 are 1,2,5,10, where the smallest one except 1 is 2, therefore the answer is 10+(f(10)=2)=12.

In the third query, n is changed as follows: 3 → 6 → 8 → 10 → 12.
/*
[ ( ^ _ ^ ) ]
*/

// problem: cf/1350/A

import java.io.*;
import java.util.*;


public class a {
    int INF = (int)1e9;
    long MOD = 1000000007;

    static int nax = 1000005;
    static int[] f;
    static void init() {
        f = new int[nax];
        f[0] = f[1] = 1;
        for(int i=2; i<nax; i++) {
            if(f[i]==0) {
                f[i] = i;
                if(1l*i*i>=nax) continue;
                for(int j=i*i; j<nax; j+=i) {
                    f[j] = i;
                }
            }
        }
    }

    void solve(InputReader in, PrintWriter out) throws IOException {
        long n = in.nextInt();
        long k = in.nextInt();
        if(n%2==0) {
            long rs = n + 2*k;
            out.println(rs);
            return;
        }
        long rs = n + (k-1)*2;
        for(int i=3; i<=n; i+=2) {
            if(n%i==0) {
                rs += i;
                break;
            }
        }
        out.println(rs);
    }
    

    public static void main(String[] args) throws IOException {
        InputReader in = new InputReader();
        PrintWriter out = new PrintWriter(System.out);
        int t = in.nextInt();
        // init();
        while(t-- >0) {
            new a().solve(in, out);
        }
        out.close();
    }
    
    public static void show(Object... o) {
        System.out.println(Arrays.deepToString(o));
    }
    
    static class InputReader {
        static BufferedReader br;
        static StringTokenizer st;
        public InputReader() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }
        String next() {
            while (st == null || !st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }
        int nextInt() { return Integer.parseInt(next()); }
        long nextLong() { return Long.parseLong(next()); }
        double nextDouble() { return Double.parseDouble(next()); }
    }
}