#include <bits/stdc++.h>
using namespace std;
struct custom_hash {
  static uint64_t splitmix64(uint64_t x) {
    x += 0x9e3779b97f4a7c15;
    x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
    x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
    return x ^ (x >> 31);
  }
  size_t operator()(pair<int, int> p) const {
    uint64_t x = p.first + p.second;
    static const uint64_t FIXED_RANDOM =
        chrono::steady_clock::now().time_since_epoch().count();
    return splitmix64(x + FIXED_RANDOM);
  }
};
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  int a, b;
  unordered_set<pair<int, int>, custom_hash> us;
  for (int i = 0; i < m; i++) {
    scanf("%d%d", &a, &b);
    a--, b--;
    us.insert({a, b});
    us.insert({b, a});
  }
  bool fnd;
  for (int i = 1; i * i <= n; i++) {
    if (n % i != 0) continue;
    fnd = true;
    for (auto x : us) {
      if (us.find({(x.first + i) % n, (x.second + i) % n}) == us.end()) {
        fnd = false;
        break;
      }
    }
    if (fnd) break;
    if (i == 1) continue;
    fnd = true;
    for (auto x : us) {
      if (us.find({(x.first + n / i) % n, (x.second + n / i) % n}) ==
          us.end()) {
        fnd = false;
        break;
      }
    }
    if (fnd) break;
  }
  if (fnd)
    puts("Yes");
  else
    puts("No");
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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 custom_hash {
  static uint64_t splitmix64(uint64_t x) {
    x += 0x9e3779b97f4a7c15;
    x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
    x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
    return x ^ (x >> 31);
  }
  size_t operator()(pair<int, int> p) const {
    uint64_t x = p.first + p.second;
    static const uint64_t FIXED_RANDOM =
        chrono::steady_clock::now().time_since_epoch().count();
    return splitmix64(x + FIXED_RANDOM);
  }
};
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  int a, b;
  unordered_set<pair<int, int>, custom_hash> us;
  for (int i = 0; i < m; i++) {
    scanf("%d%d", &a, &b);
    a--, b--;
    us.insert({a, b});
    us.insert({b, a});
  }
  bool fnd;
  for (int i = 1; i * i <= n; i++) {
    if (n % i != 0) continue;
    fnd = true;
    for (auto x : us) {
      if (us.find({(x.first + i) % n, (x.second + i) % n}) == us.end()) {
        fnd = false;
        break;
      }
    }
    if (fnd) break;
    if (i == 1) continue;
    fnd = true;
    for (auto x : us) {
      if (us.find({(x.first + n / i) % n, (x.second + n / i) % n}) ==
          us.end()) {
        fnd = false;
        break;
      }
    }
    if (fnd) break;
  }
  if (fnd)
    puts("Yes");
  else
    puts("No");
}

```