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>
# @-*- coding: utf-8 -*-
# @Time:   2018-10-21T21:21:26+08:00
# @Email:  ykaiwan@163.com


import sys
import math
import bisect
from collections import defaultdict

MOD = int(1e9+7)



# n = map(int, raw_input().split())
# input = map(int, raw_input().split())

n, m = map(int, raw_input().split())
da = set()
oor = (n, n)
zd = 0
for i in range(m):
    a, b = map(int, raw_input().split())
    da.add((a-1,b-1))

ck = set([1])

for i in range(2,int(math.sqrt(n))+1):
    if n%i==0:
        ck.add(i)
        ck.add(n/i)

for ick in ck:
    f = 1
    for a, b in da:
        if ((a+ick)%n, (b+ick)%n) not in da and ((b+ick)%n, (a+ick)%n) not in da:
            # print ick, a,b
            f=0
            break
    if f:
        # print ick
        print "YES"
        exit(0)
print "NO"
