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>
#include <bits/stdc++.h>
int dr[] = {2, 2, -2, -2, 1, -1, 1, -1};
int dc[] = {1, -1, 1, -1, 2, 2, -2, -2};
int dr1[] = {0, 0, 1, 1, 1, -1, -1, -1};
int dc1[] = {1, -1, 1, 0, -1, 0, 1, -1};
int dr2[] = {0, 0, 1, -1};
int dc2[] = {1, -1, 0, 0};
using namespace std;
map<pair<int, int>, int> mp;
int ind[200005];
int mark[20000059 + 5];
long long int pr = 1000033;
pair<int, int> v[200005];
long long int check(long long int k, long long int n, long long int m) {
  long long int i;
  long long int cn = 0;
  for (long long int i = 1; i < m + 1; i++) {
    long long int x = v[i].first;
    long long int y = v[i].second;
    long long int x1 = (x + k) % n;
    if (x1 == 0) x1 = n;
    long long int y1 = (y + k) % n;
    if (y1 == 0) y1 = n;
    if (x1 > y1) swap(x1, y1);
    if (mp.count({x1, y1}) > 0) {
      long long int p = mp[{x1, y1}];
      if (p != 0 && ind[p] != k) {
        cn++;
        ind[p] = k;
      }
    }
  }
  if (cn == m) return 1;
  return 0;
}
int main() {
  int n, m, x, y, i;
  scanf("%d%d", &n, &m);
  for (long long int i = 1; i < m + 1; i++) {
    scanf("%d%d", &v[i].first, &v[i].second);
    if (v[i].first > v[i].second) swap(v[i].first, v[i].second);
    mp[v[i]] = i;
  }
  long long int s = 0;
  bool f = 0;
  for (i = 1; i <= sqrt(n); i++) {
    if (n % i == 0) {
      long long int p = check(i, n, m);
      if (p == 1) f = 1;
      if (f == 1) break;
      if (i != 1 && i * i != n) p = check(n / i, n, m);
      if (p == 1) f = 1;
      if (f == 1) break;
    }
    if (f == 1) break;
  }
  if (f == 1)
    printf("Yes\n");
  else
    printf("No\n");
  return 0;
}
