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.
'''
    Auther: ghoshashis545 Ashis Ghosh
    College: jalpaiguri Govt Enggineerin College
    Date:12/05/2020

'''
import sys
from collections import deque,defaultdict as dd 
from bisect import bisect,bisect_left,bisect_right,insort,insort_left,insort_right
from itertools import permutations
from datetime import datetime
from math import ceil,sqrt,log,gcd
def ii():return int(input())
def si():return input()
def mi():return map(int,input().split())
def li():return list(mi())
abc='abcdefghijklmnopqrstuvwxyz'
abd={'a': 0, 'b': 1, 'c': 2, 'd': 3, 'e': 4, 'f': 5, 'g': 6, 'h': 7, 'i': 8, 'j': 9, 'k': 10, 'l': 11, 'm': 12, 'n': 13, 'o': 14, 'p': 15, 'q': 16, 'r': 17, 's': 18, 't': 19, 'u': 20, 'v': 21, 'w': 22, 'x': 23, 'y': 24, 'z': 25}
mod=1000000007
#mod=998244353
inf = float("inf")
vow=['a','e','i','o','u']
dx,dy=[-1,1,0,0],[0,0,1,-1]
def read():
    tc=0
    if tc:
        input=sys.stdin.readline
    else:
        sys.stdin=open('input1.txt', 'r')
        sys.stdout=open('output1.txt','w')





def isprime(n):
    for i in range(2,int(sqrt(n))+1):
        if n%i==0:
            return 0
    return 1
def fact():
    n=2
    while(len(p)<=200):
        if(isprime(n)):
            p.append(n)
        n+=1



def solve():

   

    for _ in range(ii()):
        n,k=mi()
        for i in range(k):
            c=-1
            for j in range(2,int(sqrt(n))+1):
                if n%j==0:
                    f=1
                    c=j
                    break
            if(c==-1):
                c=n
            if(c==2):
                n+=2*(k-i)
                break
            else:
                n+=c
        print(n)














if __name__== "__main__":
    # read()
    solve()