Educational Codeforces Round 89 (Rated for Div. 2)~~D. Two Divisors

You are given n integers a1,a2,…,an
For each aifind its two divisors d1>1 and d2>1 such that gcd(d1+d2,ai)=1 (where gcd(a,b) is the greatest common divisor of a and b) or say that there is no such pair.

Input

The first line contains single integer n(1≤n≤5⋅105) — the size of the array a. The second line contains n integers a1,a2,…,an (2≤ai≤107) — the array a.

Output

To speed up the output, print two lines with n
integers in each line.The i-th integers in the first and second lines should be corresponding divisors d1>1 and d2>1 such that gcd(d1+d2,ai)=1 or −1 and −1
if there is no such pair. If there are multiple answers, print any of them.

Example
Input

10
2 3 4 5 6 7 8 9 10 24

Output

-1 -1 -1 -1 3 -1 -1 -1 2 2
-1 -1 -1 -1 2 -1 -1 -1 5 3

Note

Let’s look at a7=8. It has 3 divisors greater than 1: 2, 4, 8. As you can see, the sum of any pair of divisors is divisible by 2 as well as a7.
There are other valid pairs of d1 and d2 for a10=24, like (3,4) or (8,3). You can print any of them.

定理1：gcd(x,y) = 1，gcd(x+y, xy) = 1
定理2：n = p1^k1+p2^k2+……
由兩個定理可知 n=p^k 輸出 -1，-1，n=n = p1^k1+p2^k2+…… 輸出 p1^k1 , n / p1^k1.
具體實現，需要用埃氏篩法把n的最小因子篩出來，進行預處理。

#include<bits/stdc++.h>

using namespace std;

typedef long long LL;
const int mod = 1e9 + 7;
const int maxn = 1e6 + 1;
const int inf = 0x3f3f3f3f;
const int M = 1e7+1;
const double Pi = acos(-1.0);
const LL INF = 0x3f3f3f3f3f3f3f3f;

template<class T, class F> inline void mem(T a,F b, int c) {for(int i=0;i<=c;++i)a[i]=b;}
template<class T> inline void read(T &x,T xk=10) { // xk 爲進制
	char ch = getchar(); T f = 1, t = 0.1;
	for(x=0; ch>'9'||ch<'0'; ch=getchar()) if(ch=='-')f=-1;
	for(;ch<='9'&&ch>='0';ch=getchar())x=x*xk+ch-'0';if(ch=='.') 
	for(ch=getchar();ch<='9'&&ch>='0';ch=getchar(),t*=0.1)x+=t*(ch-'0');x*=f;
}

int b[maxn], c[maxn];
int m[M];

int main() {
    memset(b, -1, sizeof(b));
    memset(c, -1, sizeof(c));
    for(LL i = 2; i < 10000001; ++ i) {
        if (m[i]) continue;
        m[i] = i;
        for(LL j = i * i; j < 10000001; j += i) {
            m[j] = i;
        }
    }
    int n; read(n);
    for(int i = 1; i <= n; ++ i) {
        int pre; read(pre); 
        int j = m[pre];
        int A = 1;
        while(pre % j == 0) pre /= j, A *= j;
        if (pre != 1) b[i] = A, c[i] = pre;
    }
    for(int i = 1; i <= n; ++ i) {
        printf("%d ", b[i]);
    }
    printf("\n");
    for(int j = 1; j <= n; ++ j) {
        printf("%d ", c[j]);
    }
    return 0;
}