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.awt.Point;
import java.util.*;
public class D {
	static class Pair implements Comparable<Pair> {
		long x;
		long y;
		Pair(long a, long b) {
			x = a;
			y = b;
		}
		public String toString() {
			return x + " " + y;
		}
		public int compareTo(Pair b) {
			if (this.x < b.x) return -1;
			else if (this.x > b.x) return 1;
			else if (this.y > b.y) return -1;
			else if (this.y < b.y) return 1;
			else return 0;
		}
	}
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		long x = in.nextLong();
		long b = 1;
		long r = b * (b + 1) / 2 * (2 * b + 1)/3 - b * (b + 1)/2;
		TreeSet<Pair> list = new TreeSet<>();
		while (r < x) {
			long n = b;
			long y = x + (b - 1) * b / 2 * (n + 1) - r;
			long z = b * (n + 1) - b * (b + 1) / 2;
			if (y % z == 0) {
				long m = y / z;
				list.add(new Pair(n, m));
				list.add(new Pair(m, n));
			}
			++b;
			r = b * (b + 1) / 2 * (2 * b + 1)/3 - b * (b + 1)/2;
		}
		System.out.println(list.size());
		for (Pair p : list) {
			System.out.println(p);
		}
	}
}