Inaka has a disc, the circumference of which is n units. The circumference is equally divided by n points numbered clockwise from 1 to n, such that points i and i + 1 (1 ≤ i < n) are adjacent, and so are points n and 1.

There are m straight segments on the disc, the endpoints of which are all among the aforementioned n points.

Inaka wants to know if her image is rotationally symmetrical, i.e. if there is an integer k (1 ≤ k < n), such that if all segments are rotated clockwise around the center of the circle by k units, the new image will be the same as the original one.

Input

The first line contains two space-separated integers n and m (2 ≤ n ≤ 100 000, 1 ≤ m ≤ 200 000) — the number of points and the number of segments, respectively.

The i-th of the following m lines contains two space-separated integers a_i and b_i (1 ≤ a_i, b_i ≤ n, a_i ≠ b_i) that describe a segment connecting points a_i and b_i.

It is guaranteed that no segments coincide.

Output

Output one line — "Yes" if the image is rotationally symmetrical, and "No" otherwise (both excluding quotation marks).

You can output each letter in any case (upper or lower).

Examples

Input


12 6
1 3
3 7
5 7
7 11
9 11
11 3


Output


Yes


Input


9 6
4 5
5 6
7 8
8 9
1 2
2 3


Output


Yes


Input


10 3
1 2
3 2
7 2


Output


No


Input


10 2
1 6
2 7


Output


Yes

Note

The first two examples are illustrated below. Both images become the same as their respective original ones after a clockwise rotation of 120 degrees around the center.

<image>
import java.io.*;
import java.util.*;

public class Solution {

	public static void main(String[] args) {
		Scanner in = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
		int n = in.nextInt(), m = in.nextInt();
		List<HashSet<Integer>> a = new ArrayList<HashSet<Integer>>();
		for(int i = 0; i < n; i++)
			a.add(new HashSet<Integer>());
		for(int i = 0; i < m; i++) {
			int l = in.nextInt(), r = in.nextInt();
			a.get(l - 1).add((r - l + n) % n);
			a.get(r - 1).add((l - r + n) % n);
		}
		boolean res = false;
		for(int i = 1; !res && i <= n / 2; i++){
			if(n % i != 0)
				continue;
			boolean temp = true;
			for(int j = 0; temp && j < n; j++) {
				temp &= a.get(j).equals(a.get((j + i) % n));
			}
			res |= temp;
		}
		if(res)
			System.out.println("Yes");
		else
			System.out.println("No");
		in.close();

	}

}
