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 array(dtype=int):
    return map(dtype, raw_input().split())


def gcd(a, b):
    if a % b == 0:
        return b
    return gcd(b, a % b)


def lcm(a, b):
    return a * b // gcd(a, b)


def lcm_array(A):
    if len(A) == 1:
        return A[0]

    p = 1
    for a in A:
        p = lcm(p, a)

    return p


def computeLPSArray(string, M, lps):
    length = 0  # length of the previous longest prefix suffix
    i = 1

    lps[0] = 0  # lps[0] is always 0

    # the loop calculates lps[i] for i = 1 to M-1
    while i < M:
        if string[i] == string[length]:
            length += 1
            lps[i] = length
            i += 1
        else:
            if length != 0:
                # This is tricky. Consider the example AAACAAAA
                # and i = 7.
                length = lps[length - 1]

                # Also, note that we do not increment i here
            else:
                lps[i] = 0
                i += 1


def find_period(A, N):
    n = len(A)
    lps = [0] * n

    computeLPSArray(A, n, lps)

    length = lps[n - 1]

    if length > 0 and n % (n - length) == 0:
        d = sum(A[:n - length])
        # print 'periodic'
        if sum(A) == N:
            return d
        else:
            return -1
    else:
        return -1


n, m = array()
segs = {}
for i in range(m):
    a, b = array()
    l = min((b - a + n) % n, (a - b + n) % n)
    if l not in segs:
        segs[l] = []
    if (b - a + n) % n == l:
        segs[l].append(a - 1)
    else:
        segs[l].append(b - 1)

repeating = True

# print n, segs

periods = []

for l in segs:

    A = segs[l]

    if l * 2 == n:
        A.extend([a + l for a in A])

    A.sort()
    segs[l] = [A[i] - A[i - 1] for i in range(1, len(A))]
    segs[l].append(n - sum(segs[l]))

    # print l, ':',
    # print segs[l]

    p = find_period([a for a in segs[l]], n)
    if p > 0:
        periods.append(p)
    else:
        repeating = False
        break

per = lcm_array(periods)

print 'Yes' if repeating and per < n else 'No'
