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;
#pragma GCC optimize("Ofast,no-stack-protector")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("unroll-loops")
template <class T>
inline T bigMod(T p, T e, T M) {
  T ret = 1;
  for (; e > 0; e >>= 1) {
    if (e & 1) ret = (ret * p) % M;
    p = (p * p) % M;
  }
  return (T)ret;
}
template <class T>
inline T modInverse(T a, T M) {
  return bigMod(a, M - 2, M);
}
template <class T>
inline T gcd(T a, T b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
template <class T>
inline T lcm(T a, T b) {
  a = abs(a);
  b = abs(b);
  return (a / gcd(a, b)) * b;
}
template <class T>
inline string int2String(T a) {
  ostringstream str;
  str << a;
  return str.str();
}
const int dr[] = {0, 1, 0, -1, -1, 1, 1, -1, -2, -2, 2, 2, -1, -1, 1, 1};
const int dc[] = {1, 0, -1, 0, 1, 1, -1, -1, -1, 1, -1, 1, -2, 2, -2, 2};
const int maxn = 1e5 + 5;
vector<int> v[maxn];
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  int x, y;
  for (int i = int(0); i < int(m); i++) {
    scanf("%d%d", &x, &y);
    v[x].push_back((x - y + n) % n);
    v[y].push_back((y - x + n) % n);
  }
  if (n == 2) {
    puts("Yes");
    return 0;
  }
  for (int i = int(1); i < int(n + 1); i++) sort(v[i].begin(), v[i].end());
  for (int k = 2; k * k <= n; k++) {
    if (n % k) continue;
    ;
    bool possible = 1;
    for (int i = int(1); i < int(n + 1); i++) {
      if (v[i] != v[1 + (i + k - 1 + n) % n]) {
        possible = 0;
        break;
      }
    }
    if (possible) {
      puts("Yes");
      return 0;
    }
    possible = 1;
    for (int i = int(1); i < int(n + 1); i++) {
      if (v[i] != v[1 + (i + n / k - 1 + n) % n]) {
        possible = 0;
        break;
      }
    }
    if (possible) {
      puts("Yes");
      return 0;
    }
  }
  puts("No");
  return 0;
}
