#include <bits/stdc++.h>
using namespace std;
void open(const char *s) {}
const double eps = 1e-8;
const int inf = 0x3f3f3f3f;
const int Minf = 0x7f7f7f7f;
const long long INF = 0x3f3f3f3f3f3f3f3fll;
const long long MINF = 0x7f7f7f7f7f7f7f7fll;
const int N = 1e5 + 50;
int p[2] = {123321, 432344};
int q[2] = {1232123, 235432};
unsigned int Hash[2][N * 2];
unsigned int pw[2][N * 2];
unsigned int get(int l, int r, int id) {
  return Hash[id][r] - Hash[id][l - 1] * pw[id][r - l + 1];
}
int main() {
  int n, m, x, y;
  scanf("%d%d", &n, &m);
  vector<int> v[n + 1];
  for (int i = (1); i <= (m); ++i) {
    scanf("%d%d", &x, &y);
    v[x].push_back(y - x < 0 ? y - x + n : y - x);
    v[y].push_back(x - y < 0 ? x - y + n : x - y);
  }
  for (int i = (1); i <= (n); ++i) {
    sort(v[i].begin(), v[i].end());
    for (int j = (0); j <= (1); ++j)
      for (int k = (0); k < ((int)v[i].size()); ++k) {
        Hash[j][i] = (Hash[j][i] * p[j] ^ v[i][k]) * p[j] + v[i][k];
      }
  }
  for (int i = (0); i <= (1); ++i)
    for (int j = (1); j <= (n - 1); ++j) {
      Hash[i][j + n] = Hash[i][j];
    }
  pw[0][0] = pw[1][0] = 1;
  for (int i = (0); i <= (1); ++i)
    for (int j = (1); j <= (2 * n - 1); ++j) {
      Hash[i][j] = Hash[i][j - 1] * q[i] + Hash[i][j];
      pw[i][j] = pw[i][j - 1] * q[i];
    }
  unsigned int ans1 = Hash[0][n];
  unsigned int ans2 = Hash[1][n];
  for (int i = (2); i <= (n); ++i) {
    if (ans1 == get(i, i + n - 1, 0) && ans2 == get(i, i + n - 1, 1)) {
      puts("Yes");
      return 0;
    }
  }
  puts("No");
  return 0;
}

### Prompt

Create a solution in cpp for 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;
void open(const char *s) {}
const double eps = 1e-8;
const int inf = 0x3f3f3f3f;
const int Minf = 0x7f7f7f7f;
const long long INF = 0x3f3f3f3f3f3f3f3fll;
const long long MINF = 0x7f7f7f7f7f7f7f7fll;
const int N = 1e5 + 50;
int p[2] = {123321, 432344};
int q[2] = {1232123, 235432};
unsigned int Hash[2][N * 2];
unsigned int pw[2][N * 2];
unsigned int get(int l, int r, int id) {
  return Hash[id][r] - Hash[id][l - 1] * pw[id][r - l + 1];
}
int main() {
  int n, m, x, y;
  scanf("%d%d", &n, &m);
  vector<int> v[n + 1];
  for (int i = (1); i <= (m); ++i) {
    scanf("%d%d", &x, &y);
    v[x].push_back(y - x < 0 ? y - x + n : y - x);
    v[y].push_back(x - y < 0 ? x - y + n : x - y);
  }
  for (int i = (1); i <= (n); ++i) {
    sort(v[i].begin(), v[i].end());
    for (int j = (0); j <= (1); ++j)
      for (int k = (0); k < ((int)v[i].size()); ++k) {
        Hash[j][i] = (Hash[j][i] * p[j] ^ v[i][k]) * p[j] + v[i][k];
      }
  }
  for (int i = (0); i <= (1); ++i)
    for (int j = (1); j <= (n - 1); ++j) {
      Hash[i][j + n] = Hash[i][j];
    }
  pw[0][0] = pw[1][0] = 1;
  for (int i = (0); i <= (1); ++i)
    for (int j = (1); j <= (2 * n - 1); ++j) {
      Hash[i][j] = Hash[i][j - 1] * q[i] + Hash[i][j];
      pw[i][j] = pw[i][j - 1] * q[i];
    }
  unsigned int ans1 = Hash[0][n];
  unsigned int ans2 = Hash[1][n];
  for (int i = (2); i <= (n); ++i) {
    if (ans1 == get(i, i + n - 1, 0) && ans2 == get(i, i + n - 1, 1)) {
      puts("Yes");
      return 0;
    }
  }
  puts("No");
  return 0;
}

```