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>
using namespace std;
;
const int BufferSize = 1 << 16;
char buffer[BufferSize], *head, *tail;
inline char Getchar() {
  if (head == tail) {
    int l = fread(buffer, 1, BufferSize, stdin);
    tail = (head = buffer) + l;
  }
  return *head++;
}
inline int read() {
  int x = 0, f = 1;
  char c = Getchar();
  for (; !isdigit(c); c = Getchar())
    if (c == '-') f = -1;
  for (; isdigit(c); c = Getchar()) x = x * 10 + c - '0';
  return x * f;
}
const double eps = 1e-8;
const int mod = 1e9 + 7;
const int maxn = 3e5 + 5;
const int INF = 0x3f3f3f3f;
const long long INFLL = 0x3f3f3f3f3f3f3f3f;
int n, m;
set<pair<int, int> > s;
bool check(int k) {
  for (auto &i : s) {
    pair<int, int> j = {min((i.first + k) % n, (i.second + k) % n),
                        max((i.first + k) % n, (i.second + k) % n)};
    if (s.find(j) == s.end()) {
      return false;
    }
  }
  return true;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 0, u, v; i < m; i++) {
    scanf("%d%d", &u, &v);
    u--, v--;
    s.insert(make_pair(min(u, v), max(u, v)));
  }
  for (int i = 1; i * i <= n; i++) {
    if (n % i == 0) {
      if (check(i) || (i != 1 && check(n / i))) {
        printf("YES\n");
        return 0;
      }
    }
  }
  printf("NO\n");
  return 0;
}
