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.
from sys import stdin
from collections import deque
mod = 10**9 + 7
import sys
import random
# sys.setrecursionlimit(10**6)
from queue import PriorityQueue
# def rl():
#     return [int(w) for w in stdin.readline().split()]
from bisect import bisect_right
from bisect import bisect_left
from collections import defaultdict
from math import sqrt,factorial,gcd,log2,inf,ceil
# map(int,input().split())
# # l = list(map(int,input().split()))
# from itertools import permutations
import heapq
# input = lambda: sys.stdin.readline().rstrip()
input = lambda : sys.stdin.readline().rstrip()
from sys import stdin, stdout
from heapq import heapify, heappush, heappop
from itertools import permutations
from math import factorial as f

# def ncr(x, y):
#     return f(x) // (f(y) * f(x - y))
def ncr(n, r, p):
    num = den = 1
    for i in range(r):
        num = (num * (n - i)) % p
        den = (den * (i + 1)) % p
    return (num * pow(den,
                      p - 2, p)) % p

import sys
# input = sys.stdin.readline
# LCA
# def bfs(na):
#
#     queue = [na]
#     boo[na] = True
#     level[na] = 0
#
#     while queue!=[]:
#
#         z = queue.pop(0)
#
#         for i in hash[z]:
#
#             if not boo[i]:
#
#                 queue.append(i)
#                 level[i] = level[z] + 1
#                 boo[i] = True
#                 dp[i][0] = z
#
#
#
# def prec(n):
#
#     for i in range(1,20):
#
#         for j in range(1,n+1):
#             if dp[j][i-1]!=-1:
#                 dp[j][i] = dp[dp[j][i-1]][i-1]
#
#
# def lca(u,v):
#     if level[v] < level[u]:
#         u,v = v,u
#
#     diff = level[v] - level[u]
#
#
#     for i in range(20):
#         if ((diff>>i)&1):
#             v = dp[v][i]
#
#
#     if u == v:
#         return u
#
#
#     for i in range(19,-1,-1):
#         # print(i)
#         if dp[u][i] != dp[v][i]:
#
#             u = dp[u][i]
#             v = dp[v][i]
#
#
#     return dp[u][0]
#
# dp = []
#
#
# n = int(input())
#
# for i in range(n + 10):
#
#     ka = [-1]*(20)
#     dp.append(ka)


def sqr(n):

    for i in range(2,int(sqrt(n))+1):
        if n%i == 0:
            return i

    return n



t = int(input())

for _ in range(t):


    n,k = map(int,input().split())
    z = sqr(n)
    n+=z + 2*(k-1)
    print(n)







