Mahmoud has an array *a* consisting of *n* integers. He asked Ehab to find another array *b* of the same length such that:
 -  *b* is lexicographically greater than or equal to *a*. -  *b*<sub class="lower-index">*i*</sub> ≥ 2. -  *b* is pairwise coprime: for every 1 ≤ *i* &lt; *j* ≤ *n*, *b*<sub class="lower-index">*i*</sub> and *b*<sub class="lower-index">*j*</sub> are coprime, i. e. *GCD*(*b*<sub class="lower-index">*i*</sub>, *b*<sub class="lower-index">*j*</sub>) = 1, where *GCD*(*w*, *z*) is the greatest common divisor of *w* and *z*. 
Ehab wants to choose a special array so he wants the lexicographically minimal array between all the variants. Can you find it?

An array *x* is lexicographically greater than an array *y* if there exists an index *i* such than *x*<sub class="lower-index">*i*</sub> &gt; *y*<sub class="lower-index">*i*</sub> and *x*<sub class="lower-index">*j*</sub> = *y*<sub class="lower-index">*j*</sub> for all 1 ≤ *j* &lt; *i*. An array *x* is equal to an array *y* if *x*<sub class="lower-index">*i*</sub> = *y*<sub class="lower-index">*i*</sub> for all 1 ≤ *i* ≤ *n*.
The first line contains an integer *n* (1 ≤ *n* ≤ 10<sup class="upper-index">5</sup>), the number of elements in *a* and *b*.

The second line contains *n* integers *a*<sub class="lower-index">1</sub>, *a*<sub class="lower-index">2</sub>, ..., *a*<sub class="lower-index">*n*</sub> (2 ≤ *a*<sub class="lower-index">*i*</sub> ≤ 10<sup class="upper-index">5</sup>), the elements of *a*.
Note that in the second sample, the array is already pairwise coprime so we printed it.
5
2 3 5 4 13
2 3 5 7 11 
3
10 3 7
10 3 7 
5
7 10 2 5 5
7 10 3 11 13 
7
20 9 7 6 7 9 15
20 9 7 11 13 17 19 
10
5 3 2 2 3 3 3 4 2 5
5 3 2 7 11 13 17 19 23 29 
3
3 18 2
3 19 2 
import math

N = 10**5+50
spf = [-1]*(N+1)
for i in range(N+1):
    spf[i] = i
for i in range(2, int(math.sqrt(N))+1):
    if spf[i] == i:
        for j in range(i*2, N+1, i):
            if spf[j] == j:
                spf[j] = i

def factorize(n):
    d = {}
    while n != 1:
        p = spf[n]
        if p in d:
            d[p] += 1
        else:
            d[p] = 1
        n //= p
    return d

import math

def sieve(n):
    ass = []
    is_prime = [True]*(n+1)
    is_prime[0] = False
    is_prime[1] = False

    for i in range(2, int(math.sqrt(n))+1):
        if not is_prime[i]:
            continue
        for j in range(i*2, n+1, i):
            is_prime[j] = False
    for i in range(n+1):
        if is_prime[i]:
            ass.append(i)
    return(ass)

S = sieve(N)
S.sort()

n = int(input())
A = list(map(int, input().split()))
from collections import defaultdict
D = set()
maxp = -1
import bisect
B = []
flag = False
for a in A:
    if not flag:
        d = factorize(a)
        d = list(d.items())
        d.sort()
        b = 1
        for k, v in d:
            if k not in D:
                D.add(k)
                maxp = max(maxp, k)
                b *= k
            else:
                i = bisect.bisect_right(S, maxp)
                p = S[i]
                maxp = max(maxp, p)
                b *= p
                flag = True
        B.append(b)
    else:
        i = bisect.bisect_right(S, maxp)
        p = S[i]
        maxp = max(maxp, p)
        b = p
        B.append(b)
print(*B)
