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>
def getPrimeDivisors(n):
    primes = [True] * (n+1)
    primes[0] = False
    primes[1] = False
    for i in range(2, n+1):
        if primes[i]:
            tmp = i*2
            while tmp <= n:
                primes[tmp] = False
                tmp += i
    divs = []
    for k,v in enumerate(primes):
        if v and n % k == 0:
            divs.append(k)
    return divs

def valid(div, n, s):
    inc = n // div
    for a, b in s:
        a2, b2 = a+inc, b+inc
        if a2 > n: a2 -= n
        if b2 > n: b2 -= n
        if (a2, b2) in s or (b2, a2) in s: continue
        return False
    return True

def sol(n,s):
    divs = getPrimeDivisors(n)
    for div in divs:
        if valid(div, n, s): return 'Yes'
    return 'No'

if __name__ == '__main__':
    [n,m] = [int(x) for x in input().split()]
    s = set()
    for _ in range(m):
        s.add(tuple([int(x) for x in input().split()]))
    print(sol(n,s))
