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 sys

import collections

# see tutorial at https://codeforces.com/blog/entry/66878

n, m = map(int, input().split())

# segment length -> start
segs = set()

for _ in range(m):
    start, end = map(int, input().split())
    segs.add((start-1, end-1))



def get_divisors(n):
    """

    :param n:
    :return: divisors of n excluding n itself
    """
    divisors = []
    for i in range(1, n):
        if n % i == 0:
            divisors.append(i)
    return divisors


divisors = get_divisors(n)

# try out all divisors as the total (actual) rotation period, not just the rotation period for a subset of segments
for k in divisors:
    for sega, segb in segs:
        if ((sega + k) % n, (segb + k) % n) not in segs and ((segb + k) % n, (sega + k) % n) not in segs:
            # this k doesn't work
            break
    else:
        print('Yes')
        sys.exit()

print('No')
