#include <bits/stdc++.h>
using namespace std;
const int N = (int)2e5 + 15;
const long long inf = (long long)1e14;
struct Seg {
  int l, r;
  bool operator<(const Seg& b) const { return l != b.l ? l < b.l : r < b.r; }
};
Seg s1[N], s2[N];
bool solve(const vector<int>& v2, int n, int m) {
  for (int k : v2) {
    for (int i = 1; i <= m; i++) {
      s2[i].l = (s1[i].l - 1 + k) % n + 1;
      s2[i].r = (s1[i].r - 1 + k) % n + 1;
      if (s2[i].l > s2[i].r) {
        swap(s2[i].l, s2[i].r);
      }
    }
    sort(s2 + 1, s2 + 1 + m);
    bool ok = true;
    for (int i = 1; i <= m; i++) {
      if (s1[i].l != s2[i].l || s1[i].r != s2[i].r) {
        ok = false;
        break;
      }
    }
    if (ok) {
      return true;
    }
  }
  return false;
}
int main() {
  int n, m;
  while (~scanf("%d%d", &n, &m)) {
    for (int i = 1; i <= m; i++) {
      scanf("%d%d", &s1[i].l, &s1[i].r);
      if (s1[i].l > s1[i].r) {
        swap(s1[i].l, s1[i].r);
      }
    }
    sort(s1 + 1, s1 + 1 + m);
    vector<int> v1, v2;
    for (int i = 1; i < n; i++) {
      if (n % i == 0) {
        v1.push_back(i);
      }
    }
    for (int i = 0; i < v1.size(); i++) {
      bool ok = true;
      for (int j = i + 1; j < v1.size(); j++) {
        if (v1[j] % v1[i] == 0) {
          ok = false;
          break;
        }
      }
      if (ok) {
        v2.push_back(v1[i]);
      }
    }
    puts(solve(v2, n, m) ? "Yes" : "No");
  }
}

### 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;
const int N = (int)2e5 + 15;
const long long inf = (long long)1e14;
struct Seg {
  int l, r;
  bool operator<(const Seg& b) const { return l != b.l ? l < b.l : r < b.r; }
};
Seg s1[N], s2[N];
bool solve(const vector<int>& v2, int n, int m) {
  for (int k : v2) {
    for (int i = 1; i <= m; i++) {
      s2[i].l = (s1[i].l - 1 + k) % n + 1;
      s2[i].r = (s1[i].r - 1 + k) % n + 1;
      if (s2[i].l > s2[i].r) {
        swap(s2[i].l, s2[i].r);
      }
    }
    sort(s2 + 1, s2 + 1 + m);
    bool ok = true;
    for (int i = 1; i <= m; i++) {
      if (s1[i].l != s2[i].l || s1[i].r != s2[i].r) {
        ok = false;
        break;
      }
    }
    if (ok) {
      return true;
    }
  }
  return false;
}
int main() {
  int n, m;
  while (~scanf("%d%d", &n, &m)) {
    for (int i = 1; i <= m; i++) {
      scanf("%d%d", &s1[i].l, &s1[i].r);
      if (s1[i].l > s1[i].r) {
        swap(s1[i].l, s1[i].r);
      }
    }
    sort(s1 + 1, s1 + 1 + m);
    vector<int> v1, v2;
    for (int i = 1; i < n; i++) {
      if (n % i == 0) {
        v1.push_back(i);
      }
    }
    for (int i = 0; i < v1.size(); i++) {
      bool ok = true;
      for (int j = i + 1; j < v1.size(); j++) {
        if (v1[j] % v1[i] == 0) {
          ok = false;
          break;
        }
      }
      if (ok) {
        v2.push_back(v1[i]);
      }
    }
    puts(solve(v2, n, m) ? "Yes" : "No");
  }
}

```