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.util.*;

public class Template {

	public static void main(String[] args) {
		FastScanner sc = new FastScanner();
		int yo = sc.nextInt();
		while (yo-- > 0) {
			long n = sc.nextLong();
			if(n%2 == 0) {
				System.out.println(n/2 + " " + n/2);
			}
			else {
				long a = -1;
				long b = -1;
				
				for(int i = 2; i <= 1000000; i++) {
					if(n%i == 0) {
						a = i;
						break;
					}
				}
				
				if(a == -1 && b == -1) {
					System.out.println(1 + " " + (n-1));
				}
				else {
					System.out.println(n/a + " " + (n-n/a));
				}
				
			}
		}
	}

	static class Pair {
		int x;
		int y;

		public Pair(int x, int y) {
			this.x = x;
			this.y = y;
		}
	}

	// Use this instead of Arrays.sort() on an array of ints. Arrays.sort() is n^2
	// worst case since it uses a version of quicksort. Although this would never
	// actually show up in the real world, in codeforces, people can hack, so
	// this is needed.
	static void ruffleSort(int[] a) {
		//ruffle
		int n = a.length;
		Random r = new Random();
		for (int i = 0; i < a.length; i++) {
			int oi = r.nextInt(n), temp = a[i];
			a[i] = a[oi];
			a[oi] = temp;
		}

		//then sort
		Arrays.sort(a);
	}

	static long gcd(long a, long b) {
		if (b == 0)
			return a;
		return gcd(b, a % b);
	}

	static boolean[] sieve(int n) {
		boolean isPrime[] = new boolean[n + 1];
		for (int i = 2; i <= n; i++) {
			if (isPrime[i])
				continue;
			for (int j = 2 * i; j <= n; j += i) {
				isPrime[j] = true;
			}
		}
		return isPrime;
	}

	static int mod = 1000000007;

	static long pow(int a, long b) {
		if (b == 0) {
			return 1;
		}
		if (b == 1) {
			return a;
		}
		if (b % 2 == 0) {
			long ans = pow(a, b / 2);
			return ans * ans;
		} else {
			long ans = pow(a, (b - 1) / 2);
			return a * ans * ans;
		}

	}

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

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

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

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

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

	//	For Input.txt and Output.txt	
	//	FileInputStream in = new FileInputStream("input.txt");
	//	FileOutputStream out = new FileOutputStream("output.txt");
	//	PrintWriter pw = new PrintWriter(out);
	//	Scanner sc = new Scanner(in);
}
