#include <bits/stdc++.h>
using namespace std;
const long long INF = 1e9;
long long MOD = 1e9 + 7;
int l[200005], r[200005];
int n, m;
map<pair<int, int>, int> M;
vector<int> vec[200005];
int solve(int mod, int len) {
  set<int> s;
  for (int i = 0; i < vec[len].size(); i++) {
    s.insert(vec[len][i]);
  }
  while (!s.empty()) {
    int ind = (*s.begin());
    int st = ind;
    s.erase(st);
    for (int i = ind + mod;; i += mod) {
      if (i > n) i -= n;
      if (i == st) break;
      if (s.count(i) == 0) return 0;
      s.erase(i);
    }
  }
  return 1;
}
int main() {
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    int x, y;
    scanf("%d%d", &l[i], &r[i]);
    x = l[i], y = r[i];
    if (y < x) swap(x, y);
    int l1 = (y - x), l2 = n - (y - x);
    if (l1 == l2) {
      vec[l1].push_back(x);
      vec[l1].push_back(y);
    } else if (l1 < l2) {
      vec[l1].push_back(x);
    } else {
      vec[l2].push_back(y);
    }
  }
  for (int i = n - 1; i >= 1; i--) {
    if (n % i) continue;
    int bol = 1;
    for (int j = 1; bol && j <= n / 2; j++) {
      if (solve(i, j))
        continue;
      else
        bol = 0;
    }
    if (bol) {
      cout << "Yes";
      return 0;
    }
  }
  cout << "No";
}

### 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;
const long long INF = 1e9;
long long MOD = 1e9 + 7;
int l[200005], r[200005];
int n, m;
map<pair<int, int>, int> M;
vector<int> vec[200005];
int solve(int mod, int len) {
  set<int> s;
  for (int i = 0; i < vec[len].size(); i++) {
    s.insert(vec[len][i]);
  }
  while (!s.empty()) {
    int ind = (*s.begin());
    int st = ind;
    s.erase(st);
    for (int i = ind + mod;; i += mod) {
      if (i > n) i -= n;
      if (i == st) break;
      if (s.count(i) == 0) return 0;
      s.erase(i);
    }
  }
  return 1;
}
int main() {
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    int x, y;
    scanf("%d%d", &l[i], &r[i]);
    x = l[i], y = r[i];
    if (y < x) swap(x, y);
    int l1 = (y - x), l2 = n - (y - x);
    if (l1 == l2) {
      vec[l1].push_back(x);
      vec[l1].push_back(y);
    } else if (l1 < l2) {
      vec[l1].push_back(x);
    } else {
      vec[l2].push_back(y);
    }
  }
  for (int i = n - 1; i >= 1; i--) {
    if (n % i) continue;
    int bol = 1;
    for (int j = 1; bol && j <= n / 2; j++) {
      if (solve(i, j))
        continue;
      else
        bol = 0;
    }
    if (bol) {
      cout << "Yes";
      return 0;
    }
  }
  cout << "No";
}

```