#include <bits/stdc++.h>
using namespace std;
vector<int> V[100000 + 9];
int f[400000 + 9], n, m, len = 0;
unsigned long long g[400000 + 9], ll = 1;
unsigned long long seek = 131;
void init() {
  g[0] = f[0];
  for (int i = 1; i < len * 2; i++) g[i] = g[i - 1] * seek + f[i];
}
unsigned long long getp(int p) {
  if (p == 0) {
    return g[len - 1];
  } else {
    return g[p + len - 1] - g[p - 1] * ll;
  }
}
int main() {
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    if (a > b) swap(a, b);
    if ((b - a) * 2 > n) {
      V[b].push_back(a + n - b);
    } else {
      V[a].push_back(b - a);
    }
  }
  for (int i = 1; i <= n; i++) sort(V[i].begin(), V[i].end());
  for (int i = 1; i <= n; i++) {
    for (int k = 0; k < (int)V[i].size(); k++) f[len++] = V[i][k];
  }
  if (len == 1 && f[0] * 2 == n) {
    printf("Yes");
    return 0;
  }
  for (int i = 0; i < len; i++) f[len + i] = f[i];
  init();
  for (int i = 0; i < len; i++) ll *= seek;
  unsigned long long ans = getp(0);
  for (int i = 1; i < len; i++) {
    if (getp(i) == ans) {
      printf("Yes");
      return 0;
    }
  }
  printf("No");
  return 0;
}

### Prompt

Your task is to create 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;
vector<int> V[100000 + 9];
int f[400000 + 9], n, m, len = 0;
unsigned long long g[400000 + 9], ll = 1;
unsigned long long seek = 131;
void init() {
  g[0] = f[0];
  for (int i = 1; i < len * 2; i++) g[i] = g[i - 1] * seek + f[i];
}
unsigned long long getp(int p) {
  if (p == 0) {
    return g[len - 1];
  } else {
    return g[p + len - 1] - g[p - 1] * ll;
  }
}
int main() {
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    if (a > b) swap(a, b);
    if ((b - a) * 2 > n) {
      V[b].push_back(a + n - b);
    } else {
      V[a].push_back(b - a);
    }
  }
  for (int i = 1; i <= n; i++) sort(V[i].begin(), V[i].end());
  for (int i = 1; i <= n; i++) {
    for (int k = 0; k < (int)V[i].size(); k++) f[len++] = V[i][k];
  }
  if (len == 1 && f[0] * 2 == n) {
    printf("Yes");
    return 0;
  }
  for (int i = 0; i < len; i++) f[len + i] = f[i];
  init();
  for (int i = 0; i < len; i++) ll *= seek;
  unsigned long long ans = getp(0);
  for (int i = 1; i < len; i++) {
    if (getp(i) == ans) {
      printf("Yes");
      return 0;
    }
  }
  printf("No");
  return 0;
}

```