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,math
from fractions import gcd
from bisect import bisect_left, bisect
from collections import defaultdict
from io import BytesIO
sys.stdin = BytesIO(sys.stdin.read())
input = lambda: sys.stdin.readline().rstrip('\r\n')
#n = int(input())
n,h = [int(_) for _ in input().split()]

d = defaultdict(set)
s = set()
for i in range(h):
    s,f = [int(_) for _ in input().split()]
    s,f = min(s,f),max(s,f)
    diff = min(f-s,s+n-f)
    d[diff].add((s,f))

gc = n
for k in d:
    if len(d[k]) == 1:
        if n % 2 == 1:
            print('No')
            exit()
        else:
            gc = n // 2
            break
    gc = gcd(len(d[k]),gc)
    if gc == 1:
        print('No')
        exit()
else:
    gc = n / gc
#print(gc)

dn = defaultdict(set)
for k in d:
    for s,f in d[k]:
        sn = s + gc
        fn = f + gc
        if sn != n:
            sn %= n
        if fn!= n:
            fn %= n
        sn,fn = min(sn,fn),max(sn,fn)
        dn[k].add((sn,fn))

for k in d:
    #print(k,d[k],dn[k])
    for el in d[k]:
        if el not in dn[k]:
            print('No')
            exit()
print('Yes')