Omkar and Last Class of Math

In Omkar's last class of math, he learned about the least common multiple, or LCMLCM. LCM(a,b)LCM(a,b) is the smallest positive integer xx which is divisible by both aa and bb.

Omkar, having a laudably curious mind, immediately thought of a problem involving the LCMLCM operation: given an integer nn, find positive integers aa and bb such that a+b=na+b=n and LCM(a,b)LCM(a,b) is the minimum value possible.

Can you help Omkar solve his ludicrously challenging math problem?

Input

Each test contains multiple test cases. The first line contains the number of test cases tt (1≤t≤101≤t≤10). Description of the test cases follows.

Each test case consists of a single integer nn (2≤n≤1092≤n≤109).

Output

For each test case, output two positive integers aa and bb, such that a+b=na+b=n and LCM(a,b)LCM(a,b) is the minimum possible.

Example

input

3
4
6
9

output

2 2
3 3
3 6

Note

For the first test case, the numbers we can choose are 1,31,3 or 2,22,2. LCM(1,3)=3LCM(1,3)=3 and LCM(2,2)=2LCM(2,2)=2, so we output 2 22 2.

For the second test case, the numbers we can choose are 1,51,5, 2,42,4, or 3,33,3. LCM(1,5)=5LCM(1,5)=5, LCM(2,4)=4LCM(2,4)=4, and LCM(3,3)=3LCM(3,3)=3, so we output 3 33 3.

For the third test case, LCM(3,6)=6LCM(3,6)=6. It can be shown that there are no other pairs of numbers which sum to 99 that have a lower LCMLCM.

题目大意：

给你一个n，找到两个a,b,使a+b=n,并且lcm(a,b)尽可能小。

#include<bits/stdc++.h>
using namespace std;
int t,x;
bool prim(int x) {
    if(x==2)
        return true;
    if(x<=1)
        return false;
    for(int i=2; i*i<=x; i++)
        if(x%i==0)
            return false;
    return true;
}
int gcd(int x) {
    int ans=-1;
    for(int i=2; i<x; i++)
        if(x%i==0)
        {
            ans=i;
            break;
        }
    ans=x/ans;
    return ans;
}
int main() {
    scanf("%d",&t);
    while(t--) {
        scanf("%d",&x);
        if(x%2==0)
            printf("%d %d
",x/2,x/2);
        else {
            if(prim(x))
                cout<<1<<" "<<x-1<<endl;
            else
                printf("%d %d
",gcd(x),x-gcd(x));
        }
    }
    return 0;
}