#include <bits/stdc++.h>
using namespace std;
int main() {
  long n, m, a, b, deg[200005], arr[200005], arr2[200005][2];
  for (long i = 0; i <= n; i++) {
    deg[i] = 0;
  }
  scanf("%ld %ld", &n, &m);
  for (long i = 1; i <= m; i++) {
    scanf("%ld %ld", &a, &b);
    if (a > b) {
      arr2[i][0] = a;
      arr2[i][1] = b;
    } else {
      arr2[i][0] = b;
      arr2[i][1] = a;
    }
    deg[a]++;
    deg[b]++;
  }
  int f = 1, dif = 0, ff = 1;
  for (long i = 1; i <= n; i++) {
    if (deg[i] != 1) {
      f = 0;
      break;
    }
  }
  if (f) {
    for (long i = 1; i <= n; i++) {
      dif = arr2[i][0] - arr2[i][1];
      if (dif > 1 && dif < n - 1) {
        ff = 0;
        break;
      }
    }
    if (ff)
      printf("Yes\n");
    else
      printf("No\n");
    return 0;
  }
  long k = 1, n2 = n / 2 + 1;
  for (long i = 1; i <= n2; i++) {
    if (n % i == 0) arr[k++] = i;
  }
  long ans = 0;
  for (long i = 1; i < k; i++) {
    if (deg[1] == deg[arr[i] + 1]) {
      long p = arr[i] + 1;
      ans = 1;
      for (long j = 2; j <= n2; j++) {
        ++p;
        p %= n;
        if (p == 0) {
          p = n;
        }
        if (deg[j] != deg[p]) {
          ans = 0;
          break;
        }
      }
    }
    if (ans) break;
  }
  if (ans)
    printf("Yes\n");
  else
    printf("No\n");
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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>

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
  long n, m, a, b, deg[200005], arr[200005], arr2[200005][2];
  for (long i = 0; i <= n; i++) {
    deg[i] = 0;
  }
  scanf("%ld %ld", &n, &m);
  for (long i = 1; i <= m; i++) {
    scanf("%ld %ld", &a, &b);
    if (a > b) {
      arr2[i][0] = a;
      arr2[i][1] = b;
    } else {
      arr2[i][0] = b;
      arr2[i][1] = a;
    }
    deg[a]++;
    deg[b]++;
  }
  int f = 1, dif = 0, ff = 1;
  for (long i = 1; i <= n; i++) {
    if (deg[i] != 1) {
      f = 0;
      break;
    }
  }
  if (f) {
    for (long i = 1; i <= n; i++) {
      dif = arr2[i][0] - arr2[i][1];
      if (dif > 1 && dif < n - 1) {
        ff = 0;
        break;
      }
    }
    if (ff)
      printf("Yes\n");
    else
      printf("No\n");
    return 0;
  }
  long k = 1, n2 = n / 2 + 1;
  for (long i = 1; i <= n2; i++) {
    if (n % i == 0) arr[k++] = i;
  }
  long ans = 0;
  for (long i = 1; i < k; i++) {
    if (deg[1] == deg[arr[i] + 1]) {
      long p = arr[i] + 1;
      ans = 1;
      for (long j = 2; j <= n2; j++) {
        ++p;
        p %= n;
        if (p == 0) {
          p = n;
        }
        if (deg[j] != deg[p]) {
          ans = 0;
          break;
        }
      }
    }
    if (ans) break;
  }
  if (ans)
    printf("Yes\n");
  else
    printf("No\n");
  return 0;
}

```