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

Omkar, having a laudably curious mind, immediately thought of a problem involving the LCM operation: given an integer n, find positive integers a and b such that a + b = n and 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 t (1 ≤ t ≤ 10). Description of the test cases follows.

Each test case consists of a single integer n (2 ≤ n ≤ 10^{9}).

Output

For each test case, output two positive integers a and b, such that a + b = n and 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, 3 or 2, 2. LCM(1, 3) = 3 and LCM(2, 2) = 2, so we output 2 \ 2.

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

For the third test case, LCM(3, 6) = 6. It can be shown that there are no other pairs of numbers which sum to 9 that have a lower LCM.
import java.io.*;
import java.math.*;
import java.util.*;

public class Sample
{

	static int[][][] dp;
	static int[] are;

	public static void main(String[] args) throws InterruptedException {
		Thread t=new Thread(null, null,"", 1<<28) {
			public void run() {
				m();
			}
		};
		t.start();
		t.join();
	}

	public static void m()
	{
		FastScanner in=new FastScanner();
		int t = in.nextInt();
		while(t-->0)
		{
			int n = in.nextInt();
			int ans = 1;
			for(int i=2; i*i<=n; i++)
			{
				if((n%i)==0)
				{
					ans=n/i;
					break;
				}
			}
			System.out.println(ans+" "+(n-ans));
		}
	}

/*
1
5
01101
10101

*/

	static void longestsub(int[] str1, int[] str2)
	{
		int n = str1.length;
		int m = str2.length;
		//
		int[][] dp = new int[n+1][m+1];
		//
		for(int i=0; i<=n; i++)
		{
			for(int j=0; j<=m; j++)
			{
				if(i==0 || j==0) {dp[i][j]=0; continue;}
				//
				if(str1[i-1]==str2[j-1]) dp[i][j]=dp[i-1][j-1]+1;
				else dp[i][j] = Math.max(dp[i-1][j], dp[i][j-1]);
			}
		}
		//
		int length = dp[n][m];
		//
		String ans = "";
		//
		int i=n, j=m;
		while(true)
		{
			if(i==0 || j==0) break;
			//
			if(dp[i][j]==Math.max(dp[i-1][j], dp[i][j-1]))
			{
				if(dp[i-1][j]>dp[i][j-1]) i--;
				else j--;
			}
			else
			{
				ans = str1[i-1] + " " + ans;
				i--; j--;
			}
		}
		//
		System.out.println(length>0?"YES"+'\n'+"1 "+ans.substring(0,1):"NO");
	}

	static void print(String[] are)
	{
		for(int i=0; i<are.length; i++)
			System.out.print(are[i]);
		System.out.println();
	}

	static void sort(int[] a)
	{
		ArrayList<Integer> list=new ArrayList<>();
		for (int i:a) list.add(i);
		Collections.sort(list);
		for (int i=0; i<a.length; i++) a[i]=list.get(i);
	}

	static class FastScanner
	{
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");

		public String next() {
			while (!st.hasMoreTokens())
				try {
					st=new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();

		}

		public int nextInt() {
			return Integer.parseInt(next());
		}

		public int[] readArray(int n) {
			int[] a=new int[n];
			for (int i=0; i<n; i++) a[i]=nextInt();
			return a;
		}

		public long nextLong() {
			return Long.parseLong(next());
		}

		public long[] readLongArray(int n) {
			long[] a=new long[n];
			for (int i=0; i<n; i++) a[i]=nextLong();
			return a;
		}
	}

}
