You are given a positive integer x. Find any such 2 positive integers a and b such that GCD(a,b)+LCM(a,b)=x.

As a reminder, GCD(a,b) is the greatest integer that divides both a and b. Similarly, LCM(a,b) is the smallest integer such that both a and b divide it.

It's guaranteed that the solution always exists. If there are several such pairs (a, b), you can output any of them.

Input

The first line contains a single integer t (1 ≤ t ≤ 100) — the number of testcases.

Each testcase consists of one line containing a single integer, x (2 ≤ x ≤ 10^9).

Output

For each testcase, output a pair of positive integers a and b (1 ≤ a, b ≤ 10^9) such that GCD(a,b)+LCM(a,b)=x. It's guaranteed that the solution always exists. If there are several such pairs (a, b), you can output any of them.

Example

Input


2
2
14


Output


1 1
6 4

Note

In the first testcase of the sample, GCD(1,1)+LCM(1,1)=1+1=2.

In the second testcase of the sample, GCD(6,4)+LCM(6,4)=2+12=14.
import java.util.Scanner;

public class EhAbAnDgCd{
    private static Scanner s = new Scanner(System.in);
    public static void main(String[] args){
        int T = s.nextInt();
        for(int z=0;z<T;z++){
            int num = s.nextInt();
            for(int i=1;i<=num/2;i++){
                if( hcf(i,num-i) + lcm(i,num-i) == num ){
                    System.out.println(i+" "+(num-i)); break;
                }
            }
        }
    }
    public static int hcf(int x, int y){
        int num = 1;
        int min = Math.min(x,y);
        for(int i=2;i<=min;i++){
            if( x%i == 0 && y%i == 0){
                num = i;
            }
        }
        return num;
    }
    public static int lcm(int x,int y){
        int max = Math.max(x,y);
        while(true){
            if(max%x == 0 && max%y == 0 ){
                return max;
            }
            max++;
        }
    }
}