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 Main {
	public static void main(String[] args) {
		Main main=new Main();
		Scanner s=new Scanner(System.in);
		while(s.hasNext()) {
			int t=s.nextInt();
			long[][] ans=new long[t][2];
			for(int i=0;i<t;i++) {
				long n=s.nextLong();
				ans[i]=main.function(n);
			}
			for(long res[]:ans) {
				for(long a:res) {System.out.print(a+" ");}
				System.out.println();
			}
		}
	}
	public long[] function(long n) {
		long x=0,y=0;
		for(int i=1;i<=n;i++) {
			x=gcd(i,n-i);
			y=lcm(i,n-i);
			if(x+y==n) {
				break;
			}
		}
		return new long[] {x,y};
	}
	
	//算最大公约数
	private static long gcd(long a, long b) {
		return b == 0 ? a : gcd(b, a % b);
    }
	
	private static long lcm(long a, long b) {
		return a*b/gcd(a,b);
    }
}
 			 		 			  	    			    			