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.
# -*- coding: utf-8 -*-
import sys
import math
import os
import itertools
import string
import heapq
import _collections
from collections import Counter
from collections import defaultdict
from functools import lru_cache
import bisect
import re
import queue
from decimal import *


class Scanner():
    @staticmethod
    def int():
        return int(sys.stdin.readline().rstrip())

    @staticmethod
    def string():
        return sys.stdin.readline().rstrip()

    @staticmethod
    def map_int():
        return [int(x) for x in Scanner.string().split()]

    @staticmethod
    def string_list(n):
        return [Scanner.string() for i in range(n)]

    @staticmethod
    def int_list_list(n):
        return [Scanner.map_int() for i in range(n)]

    @staticmethod
    def int_cols_list(n):
        return [Scanner.int() for i in range(n)]


class Math():
    @staticmethod
    def gcd(a, b):
        if b == 0:
            return a
        return Math.gcd(b, a % b)

    @staticmethod
    def lcm(a, b):
        return (a * b) // Math.gcd(a, b)

    @staticmethod
    def divisor(n):
        res = []
        i = 1
        for i in range(1, int(n ** 0.5) + 1):
            if n % i == 0:
                res.append(i)
                if i != n // i:
                    res.append(n // i)
        return res

    @staticmethod
    def round_up(a, b):
        return -(-a // b)

    @staticmethod
    def is_prime(n):
        if n < 2:
            return False
        if n == 2:
            return True
        if n % 2 == 0:
            return False
        d = int(n ** 0.5) + 1
        for i in range(3, d + 1, 2):
            if n % i == 0:
                return False
        return True


def pop_count(x):
    x = x - ((x >> 1) & 0x5555555555555555)
    x = (x & 0x3333333333333333) + ((x >> 2) & 0x3333333333333333)
    x = (x + (x >> 4)) & 0x0f0f0f0f0f0f0f0f
    x = x + (x >> 8)
    x = x + (x >> 16)
    x = x + (x >> 32)
    return x & 0x0000007f


MOD = int(1e09) + 7
INF = int(1e15)


def solve():
    N, K = Scanner.map_int()
    if N % 2 == 0:
        print(2 * K + N)
    else:
        print(2 * (K - 1) + N + sorted(Math.divisor(N))[1])


def main():
    # sys.stdin = open("sample.txt")
    T = Scanner.int()
    for _ in range(T):
        solve()


if __name__ == "__main__":
    main()
