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 M = 1e5 + 5;
const int inf = 1e9 + 5;
const int mod = 1e9 + 7;
vector<int> v[M];
int hx2[2 * M];
int hx[M];
int nxt[M];
int n, m, len;
void getNext() {
  int i = 0;
  nxt[i] = -1;
  int j = -1;
  while (i < len) {
    if (j == -1 || hx[i] == hx[j]) {
      ++i;
      ++j;
      if (hx[i] != hx[j])
        nxt[i] = j;
      else
        nxt[i] = nxt[j];
    } else
      j = nxt[j];
  }
}
int kmp_match() {
  int i = 0, j = 0;
  int cnt = 0;
  while (i <= m - 1) {
    if (j < 0 || hx2[i] == hx[j]) {
      i++;
      j++;
    } else {
      j = nxt[j];
    }
    if (j == n) {
      cnt++;
      std::cout << "["
                << "i"
                << "->" << i << "]" << endl;
      std::cout << "["
                << "j"
                << "->" << j << "]" << endl;
      j = nxt[j];
    }
  }
  return cnt;
}
int KMP() {
  memset(nxt, 0, sizeof(nxt));
  int la, lb;
  la = n;
  lb = m;
  int j, k;
  j = 0;
  k = -1;
  nxt[0] = -1;
  while (j < la) {
    if (k == -1 || hx[j] == hx[k]) {
      j++;
      k++;
      nxt[j] = k;
    } else {
      k = nxt[k];
    }
  }
  int ans = 0;
  j = 0;
  k = 0;
  while (k < lb) {
    if (j == -1 || hx[j] == hx2[k]) {
      j++;
      k++;
    } else {
      j = nxt[j];
    }
    if (j == la) {
      j = nxt[j];
      ans++;
    }
  }
  return ans;
}
int main() {
  int k;
  scanf("%d%d", &n, &k);
  m = 2 * n;
  for (int i = 0; i < k; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    if (a > b) swap(a, b);
    v[a].push_back(min(b - a, n - b + a));
    v[b].push_back(min(b - a, n - b + a));
  }
  for (int i = 0; i < n; i++) {
    sort(v[i + 1].begin(), v[i + 1].end());
    for (int j = 0; j < v[i + 1].size(); j++) {
      hx[i] = (hx[i] * n % mod + v[i + 1][j]) % mod;
    }
  }
  for (int i = 0; i < n; i++) {
    hx2[i] = hx[i];
    hx2[i + n] = hx[i];
  }
  if (KMP() <= 2) {
    printf("No\n");
  } else {
    printf("Yes\n");
  }
  return 0;
}
