#include <bits/stdc++.h>
using namespace std;
inline int in() {
  int res = 0;
  char c;
  int f = 1;
  while ((c = getchar()) < '0' || c > '9')
    if (c == '-') f = -1;
  while (c >= '0' && c <= '9') res = res * 10 + c - '0', c = getchar();
  return res * f;
}
const int N = 2e5 + 7;
const int M = 5e6 + 7;
int a[M];
int Next[M];
vector<int> v[N];
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 0; i < m; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    x--, y--;
    if (y < x) {
      v[x].push_back(y + n);
      v[y].push_back(x);
    } else {
      v[x].push_back(y);
      v[y].push_back(x + n);
    }
  }
  int cnt = 0;
  for (int i = 0; i < n; i++) {
    int sz = v[i].size();
    sort(v[i].begin(), v[i].end());
    for (int j = 0; j < sz; j++) {
      a[cnt] = v[i][j] - i;
      cnt++;
    }
    a[cnt] = -1;
    cnt++;
  }
  Next[0] = -1;
  int k = -1, j = 0;
  while (j < cnt) {
    if (k == -1 || a[j] == a[k])
      Next[++j] = ++k;
    else
      k = Next[k];
  }
  j = cnt - Next[cnt];
  if (cnt % j == 0 && j != cnt)
    printf("Yes\n");
  else
    puts("No");
  return 0;
}

### Prompt

Your challenge is to write 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;
inline int in() {
  int res = 0;
  char c;
  int f = 1;
  while ((c = getchar()) < '0' || c > '9')
    if (c == '-') f = -1;
  while (c >= '0' && c <= '9') res = res * 10 + c - '0', c = getchar();
  return res * f;
}
const int N = 2e5 + 7;
const int M = 5e6 + 7;
int a[M];
int Next[M];
vector<int> v[N];
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 0; i < m; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    x--, y--;
    if (y < x) {
      v[x].push_back(y + n);
      v[y].push_back(x);
    } else {
      v[x].push_back(y);
      v[y].push_back(x + n);
    }
  }
  int cnt = 0;
  for (int i = 0; i < n; i++) {
    int sz = v[i].size();
    sort(v[i].begin(), v[i].end());
    for (int j = 0; j < sz; j++) {
      a[cnt] = v[i][j] - i;
      cnt++;
    }
    a[cnt] = -1;
    cnt++;
  }
  Next[0] = -1;
  int k = -1, j = 0;
  while (j < cnt) {
    if (k == -1 || a[j] == a[k])
      Next[++j] = ++k;
    else
      k = Next[k];
  }
  j = cnt - Next[cnt];
  if (cnt % j == 0 && j != cnt)
    printf("Yes\n");
  else
    puts("No");
  return 0;
}

```