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 io, os
input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline

def main():

    n, m = map(int, input().split())
    S = set()
    for i in range(m):
        a, b = map(int, input().split())
        a, b = a-1, b-1
        if a > b:
            a, b = b, a
        S.add((a, b))


    def make_divisors(n):
        divisors = []
        for i in range(1, int(n**0.5)+1):
            if n % i == 0:
                divisors.append(i)
                if i != n // i:
                    divisors.append(n//i)

        #divisors.sort(reverse=True)
        return divisors
    D = make_divisors(n)
    for d in D:
        if d == n:
            continue
        T = set()
        flag = True
        for a, b in S:
            na = (a+d)%n
            nb = (b+d)%n
            if na > nb:
                na, nb = nb, na
            if (na, nb) not in S:
                flag = False
                break
            T.add((na, nb))
        if not flag:
            continue
        for a, b in S:
            if (a, b) not in T:
                break
        else:
            print('Yes')
            exit()
    else:
        print('No')

if __name__ == '__main__':
    main()
