#include <bits/stdc++.h>
using namespace std;
const long long INF = 1e9 + 7;
const int N = 2e5 + 10;
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  vector<pair<int, int>> vp;
  while (m--) {
    int x, y;
    scanf("%d%d", &x, &y);
    if (x > y) swap(x, y);
    vp.emplace_back(x, y);
  }
  sort(vp.begin(), vp.end());
  set<int> s;
  for (int i = 2; i * i <= n; i++) {
    if (n % i) continue;
    s.emplace(i);
    s.emplace(n / i);
  }
  s.emplace(1);
  while (!s.empty()) {
    int d = *s.rbegin();
    s.erase(d);
    vector<pair<int, int>> v;
    for (auto& p : vp) {
      int x, y;
      tie(x, y) = p;
      x += d;
      y += d;
      if (x > n) x -= n;
      if (y > n) y -= n;
      if (x > y) swap(x, y);
      v.emplace_back(x, y);
    }
    sort(v.begin(), v.end());
    if (v == vp) puts("Yes"), exit(0);
    int cur = 1;
    while (1) {
      auto iter = s.upper_bound(cur);
      if (iter == s.end()) break;
      cur = *iter;
      if (d % cur == 0) s.erase(cur);
    }
  }
  puts("No");
  return 0;
}

### Prompt

Please create a solution in cpp 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 long long INF = 1e9 + 7;
const int N = 2e5 + 10;
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  vector<pair<int, int>> vp;
  while (m--) {
    int x, y;
    scanf("%d%d", &x, &y);
    if (x > y) swap(x, y);
    vp.emplace_back(x, y);
  }
  sort(vp.begin(), vp.end());
  set<int> s;
  for (int i = 2; i * i <= n; i++) {
    if (n % i) continue;
    s.emplace(i);
    s.emplace(n / i);
  }
  s.emplace(1);
  while (!s.empty()) {
    int d = *s.rbegin();
    s.erase(d);
    vector<pair<int, int>> v;
    for (auto& p : vp) {
      int x, y;
      tie(x, y) = p;
      x += d;
      y += d;
      if (x > n) x -= n;
      if (y > n) y -= n;
      if (x > y) swap(x, y);
      v.emplace_back(x, y);
    }
    sort(v.begin(), v.end());
    if (v == vp) puts("Yes"), exit(0);
    int cur = 1;
    while (1) {
      auto iter = s.upper_bound(cur);
      if (iter == s.end()) break;
      cur = *iter;
      if (d % cur == 0) s.erase(cur);
    }
  }
  puts("No");
  return 0;
}

```