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
Copy
5
2 3 5 4 13
Copy
2 3 5 7 11
Copy
3
10 3 7
Copy
10 3 7
Note
Note that in the second sample, the array is already pairwise coprime so we printed it.
求一个互素的序列且字典序比 A 序列大的最小序列;
显然,当某一位置的 b[ i ]> a[ i ] 时,我们只需要安排剩下的数使得 b 数列互素即可;
那么这个我们可以用 fg 来标记一下;
如何保证我们的序列互素呢?考虑质因子,
我们用 use 来看最小质因子是否有使用,如果已经使用,那么显然不能选用;
#include<bits/stdc++.h>
using namespace std;
#define maxn 200005
int prime[2000004],pre[2000004];
int n;
int a[maxn];
int b[maxn];
bool fg,vis[2000004],use[2000005];
bool judge(int x)
{
// 检验 x 是否有已经用过的质因子
int num[60],cnt=0;
while(vis[x]){
num[++cnt]=pre[x];x/=pre[x];
}
num[++cnt]=x;
for(int i=1;i<=cnt;i++){
if(use[num[i]])return false;
}
return true;
}
int main()
{
ios::sync_with_stdio(0);
cin>>n;
for(int i=1;i<=n;i++)cin>>a[i];
int cnt=0;
for(int i=2;i<=2000000;i++){
if(!vis[i]){
prime[++cnt]=i;
}
for(int j=1;j<=cnt;j++){
int tmp=i*prime[j];
if(tmp>2000000)break;
vis[tmp]=1;
pre[tmp]=prime[j];// 最小质因子
if(i%prime[j]==0)break;
}
}
int j=1;
for(int i=1;i<=n;i++){
if(fg){
while(use[prime[j]])j++;
b[i]=prime[j];
use[prime[j]]=1;
}
else{
int tmp=a[i];
while(!judge(tmp))tmp++;
if(tmp>a[i])fg=1;
b[i]=tmp;
while(vis[tmp]){
use[pre[tmp]]=1;tmp/=pre[tmp];
}
use[tmp]=1;
}
}
for(int i=1;i<=n;i++)cout<<b[i]<<' ';
cout<<endl;
}