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>
nz = [int(x) for x in input().split()]

edges = set()
divisors = set()

for x in range (0,nz[1]):	
	uv = [ y for y in input().split()]
	edges.add(uv[0] + ' ' + uv[1])

for x in range(1,nz[0]):
	if nz[0] % x == 0:
		divisors.add(x)


flag = 0

for y in divisors:
	flag = 0
	for x in edges:
		u = (int(x.split()[0]) + y) % nz[0]
		v = (int(x.split()[1]) + y) % nz[0]

		if u == 0:
			u = nz[0]

		if v == 0:
			v = nz[0] 

		if str(u) + ' ' + str(v) not in edges:
			if str(v) + ' ' + str(u) not in edges:
				flag = 1
				break
	if flag == 0:
		print("Yes")
		break

if flag == 1:
	print("No")

	
