D. Mahmoud and Ehab and another array construction task（因子标记）

D. Mahmoud and Ehab and another array construction task

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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.
• bi ≥ 2.
• b is pairwise coprime: for every 1 ≤ i < j ≤ n, bi and bj are coprime, i. e. GCD(bi, bj) = 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 xi > yi and xj = yj for all 1 ≤ j < i. An array x is equal to an array y if xi = yi for all 1 ≤ i ≤ n.

Input

The first line contains an integer n (1 ≤ n ≤ 105), the number of elements in a and b.

The second line contains n integers a1, a2, ..., an (2 ≤ ai ≤ 105), the elements of a.

Output

Output n space-separated integers, the i-th of them representing bi.

Examples

input

Copy

5
2 3 5 4 13

output

Copy

2 3 5 7 11 

input

Copy

3
10 3 7

output

Copy

10 3 7 

Note

Note that in the second sample, the array is already pairwise coprime so we printed it.

题意是说：给定a 数组，让你找b数组，使得b满足b 的字典序恰能大于a，b的最小值大于1

且b数组的任何元素都两两互质。

思路 ： 对于b中每一个数如果选中，我们标记他所有因子的倍数这样来去重，然后就可以沿着a的边界去找b。从0开始，如果a[I]没有标记b[i] = a[i] ; 否则b[i] 取第一个没被标记大于a[i]的数。并且以后取最小的没被标记的数。

 

#include <bits/stdc++.h>
using namespace std;

bool mk[10000006];
int fact[10000006];

void get()
{
	int i = 1;
	for(i = 2; i*i <= 10000000; i++)
	{
		if(fact[i] == 0)
		{
			fact[i] = i;
			for(int j = i*i; j < 10000000; j+=i)
			if(fact[j] == 0)
				fact[j] = i;
		}
	}
	while(i <= 10000000)
	{
	    if(fact[i] == 0) fact[i] = i;
	    i++;
	}
}

void sign(int x)
{
	int a = fact[x];
	while(x%a == 0) x/= a;
	
	for(int i = a; i <= 10000000; i+=a)
		mk[i] = 1;
		
	if(x != 1) sign(x);
}

int main()
{
	int n;
	get();
	scanf("%d",&n);
	int yes = 0;
	int k = 2;
	for(int i = 0; i < n; i++)
	{
		int x;
		scanf("%d",&x);
		if(yes)
		{
			while(mk[k]) k++;
			printf("%d ",k);
			sign(k);
			continue;
		}
		if(mk[x])
		{
			while(mk[x]) x++;
			printf("%d ",x);
			sign(x);
			yes = 1;
			continue;
		}
		printf("%d ",x);
		sign(x);
	}
	return 0;
}