Problem: Spongebob is already tired trying to reason his weird actions and calculations, so he simply asked you to find all pairs of n and m, such that there are exactly x distinct squares in the table consisting of n rows and m columns. For example, in a 3 × 5 table there are 15 squares with side one, 8 squares with side two and 3 squares with side three. The total number of distinct squares in a 3 × 5 table is 15 + 8 + 3 = 26.

Input

The first line of the input contains a single integer x (1 ≤ x ≤ 1018) — the number of squares inside the tables Spongebob is interested in.

Output

First print a single integer k — the number of tables with exactly x distinct squares inside.

Then print k pairs of integers describing the tables. Print the pairs in the order of increasing n, and in case of equality — in the order of increasing m.

Examples

Input

26


Output

6
1 26
2 9
3 5
5 3
9 2
26 1


Input

2


Output

2
1 2
2 1


Input

8


Output

4
1 8
2 3
3 2
8 1

Note

In a 1 × 2 table there are 2 1 × 1 squares. So, 2 distinct squares in total.

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In a 2 × 3 table there are 6 1 × 1 squares and 2 2 × 2 squares. That is equal to 8 squares in total.

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Solution (in JAVA):

Solution:
import java.util.ArrayList;
import java.util.Scanner;
public class d {
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		long n = in.nextLong();
		in.close();
		
		ArrayList<Pair> solns = new ArrayList<Pair>();
		solns.add(new Pair(1, n));
		long sub = 0;
		for(long i = 2; i*(i+1)*(2*i+1)/6 <= n; i++) {
			long mult = i * (i+1)/2;
			if(sub == 0)
				sub = 1;
			else
				sub += i*(i-1)/2;
			
			//k * mult - sub = n
			if((n + sub)%mult == 0)
				solns.add(new Pair(i, (n+sub)/mult));
		}
		
		for(int i = solns.size()-1; i >= 0; i--) {
			Pair p = solns.get(i);
			if(p.first != p.second)
				solns.add(new Pair(p.second, p.first));
		}
		System.out.println(solns.size());
		for(Pair p : solns)
			System.out.printf("%d %d\n", p.first, p.second);
		}
	
	static class Pair {
		long first, second;
		
		public Pair(long f, long s) {
			first = f; second = s;
		}
	}
}