#include <bits/stdc++.h>
using namespace std;
const int maxn = 201010;
vector<int> v[201010];
int a[maxn], b[maxn];
int x, y, n, m;
int sum(int x, int y) {
  if (x > y) {
    return min(x - y, y - x + n);
  } else {
    return min(y - x, x - y + n);
  }
}
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= m; i++) {
    scanf("%d %d", &x, &y);
    v[x].push_back(y);
    v[y].push_back(x);
  }
  int ret = 1;
  for (int i = 1; i < n; i++) {
    if (n % i == 0) {
      int flag = 1;
      for (int j = 1; j <= n; j++) {
        if (v[j].size() != v[(j + i - 1) % n + 1].size()) {
          flag = 0;
          break;
        }
        int l = v[j].size();
        for (int k = 0; k < l; k++) {
          a[k] = sum(v[j][k], j);
          b[k] = sum(v[(j + i - 1) % n + 1][k], (j + i - 1) % n + 1);
        }
        sort(a, a + v[j].size());
        sort(b, b + v[j].size());
        for (int k = 0; k < l; k++) {
          if (a[k] != b[k]) {
            flag = 0;
            break;
          }
        }
      }
      if (flag) {
        printf("Yes\n");
        ret = 0;
        break;
      }
    }
  }
  if (ret) printf("No\n");
  return 0;
}

### Prompt

Generate a CPP solution to the following problem:

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;
const int maxn = 201010;
vector<int> v[201010];
int a[maxn], b[maxn];
int x, y, n, m;
int sum(int x, int y) {
  if (x > y) {
    return min(x - y, y - x + n);
  } else {
    return min(y - x, x - y + n);
  }
}
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= m; i++) {
    scanf("%d %d", &x, &y);
    v[x].push_back(y);
    v[y].push_back(x);
  }
  int ret = 1;
  for (int i = 1; i < n; i++) {
    if (n % i == 0) {
      int flag = 1;
      for (int j = 1; j <= n; j++) {
        if (v[j].size() != v[(j + i - 1) % n + 1].size()) {
          flag = 0;
          break;
        }
        int l = v[j].size();
        for (int k = 0; k < l; k++) {
          a[k] = sum(v[j][k], j);
          b[k] = sum(v[(j + i - 1) % n + 1][k], (j + i - 1) % n + 1);
        }
        sort(a, a + v[j].size());
        sort(b, b + v[j].size());
        for (int k = 0; k < l; k++) {
          if (a[k] != b[k]) {
            flag = 0;
            break;
          }
        }
      }
      if (flag) {
        printf("Yes\n");
        ret = 0;
        break;
      }
    }
  }
  if (ret) printf("No\n");
  return 0;
}

```