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.io.BufferedReader;
import java.io.File;
import java.io.FileInputStream;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Stack;

public class D {
	public static void main(String[] args) throws IOException {
		File inputFile = new File("entradaD");
		if (inputFile.exists())
			System.setIn(new FileInputStream(inputFile));
		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
		String line;
		StringBuilder out = new StringBuilder();
		while ((line = in.readLine()) != null) {
			long x = Long.parseLong(line.trim());
			long count = 1, a = 0, b = 0, c = 0, k = 0, m = 0;
			Stack<Object[]> s = new Stack<>();
			if (1 != x)
				s.push(new Object[] { x, 1 });
			StringBuilder out2 = new StringBuilder();
			out2.append("1 " + x + "\n");
			for (long n = 2;; n++) {
				k = n - 1;
				a = k + 1;
				b = (k * (k + 1)) / 2;
				c = (k * (k + 1) * (2 * k + 1)) / 6;
				if (a * n - b >= 0 && (x + b * n - c) < n * (a * n - b))
					break;
				if ((x + b * n - c) % (a * n - b) == 0L) {
					m = (x + b * n - c) / (a * n - b);
					count++;
					if (n < m) {
						out2.append(n + " " + m + "\n");
						s.push(new Object[] { m, n });
					} else if (n == m)
						out2.append(n + " " + m + "\n");
					else {
						count--;
						break;
					}
				}
			}
			Object[] cur;
			while (!s.isEmpty()) {
				cur = s.pop();
				out2.append(cur[0] + " " + cur[1] + "\n");
				count++;
			}
			out.append(count + "\n");
			out.append(out2.toString());
		}
		System.out.print(out);
	}
}