#include <bits/stdc++.h>
using namespace std;
struct triplet {
  int index, a, b;
};
bool comp(triplet &t1, triplet &t2) { return (t1.a < t2.a); }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, m, a, b;
  cin >> n >> m;
  vector<pair<long long int, long long int> > v1, v2;
  for (int i = 0; i < m; i++) {
    cin >> a >> b;
    v1.push_back({min(a, b), max(a, b)});
  }
  sort(v1.begin(), v1.end());
  for (int j = 0; j < m; j++) {
    a = v1[j].first;
    b = v1[j].second;
    a = a + 1;
    b = b + 1;
    if (a > n) a = a - n;
    if (b > n) b = b - n;
    v2.push_back({min(a, b), max(a, b)});
  }
  sort(v2.begin(), v2.end());
  if (v2 == v1) {
    cout << "Yes";
    return 0;
  }
  for (int i = 2; i * i <= n; i++) {
    if ((n % i) != 0) continue;
    v2.clear();
    for (int j = 0; j < m; j++) {
      a = v1[j].first;
      b = v1[j].second;
      a = a + i;
      b = b + i;
      if (a > n) a = a - n;
      if (b > n) b = b - n;
      v2.push_back({min(a, b), max(a, b)});
    }
    sort(v2.begin(), v2.end());
    if (v2 == v1) {
      cout << "Yes";
      return 0;
    }
    v2.clear();
    for (int j = 0; j < m; j++) {
      a = v1[j].first;
      b = v1[j].second;
      a = a + n / i;
      b = b + n / i;
      if (a > n) a = a - n;
      if (b > n) b = b - n;
      v2.push_back({min(a, b), max(a, b)});
    }
    sort(v2.begin(), v2.end());
    if (v2 == v1) {
      cout << "Yes";
      return 0;
    }
  }
  cout << "No";
}

### Prompt

Please formulate 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;
struct triplet {
  int index, a, b;
};
bool comp(triplet &t1, triplet &t2) { return (t1.a < t2.a); }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, m, a, b;
  cin >> n >> m;
  vector<pair<long long int, long long int> > v1, v2;
  for (int i = 0; i < m; i++) {
    cin >> a >> b;
    v1.push_back({min(a, b), max(a, b)});
  }
  sort(v1.begin(), v1.end());
  for (int j = 0; j < m; j++) {
    a = v1[j].first;
    b = v1[j].second;
    a = a + 1;
    b = b + 1;
    if (a > n) a = a - n;
    if (b > n) b = b - n;
    v2.push_back({min(a, b), max(a, b)});
  }
  sort(v2.begin(), v2.end());
  if (v2 == v1) {
    cout << "Yes";
    return 0;
  }
  for (int i = 2; i * i <= n; i++) {
    if ((n % i) != 0) continue;
    v2.clear();
    for (int j = 0; j < m; j++) {
      a = v1[j].first;
      b = v1[j].second;
      a = a + i;
      b = b + i;
      if (a > n) a = a - n;
      if (b > n) b = b - n;
      v2.push_back({min(a, b), max(a, b)});
    }
    sort(v2.begin(), v2.end());
    if (v2 == v1) {
      cout << "Yes";
      return 0;
    }
    v2.clear();
    for (int j = 0; j < m; j++) {
      a = v1[j].first;
      b = v1[j].second;
      a = a + n / i;
      b = b + n / i;
      if (a > n) a = a - n;
      if (b > n) b = b - n;
      v2.push_back({min(a, b), max(a, b)});
    }
    sort(v2.begin(), v2.end());
    if (v2 == v1) {
      cout << "Yes";
      return 0;
    }
  }
  cout << "No";
}

```