#include <bits/stdc++.h>
using namespace std;
long long n, m;
long long z[200005];
set<long long> s[200005];
long long check(long long a, long long b) {
  if (s[a].size() == s[b].size()) {
    auto itr1 = s[a].begin(), itr2 = s[b].begin();
    while (itr1 != s[a].end()) {
      if ((*itr1) != (*itr2)) return 0;
      itr1++;
      itr2++;
    }
    return 1;
  }
  return 0;
}
long long z_function() {
  n = 2 * n;
  for (long long i = 1, l = 0, r = 0; i < n; ++i) {
    if (i <= r) z[i] = min(r - i + 1, z[i - l]);
    while (i + z[i] < n && check(z[i], i + z[i])) ++z[i];
    if (i + z[i] - 1 > r) l = i, r = i + z[i] - 1;
  }
  return 0;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> m;
  for (long long i = 0; i < m; i++) {
    long long l, r;
    cin >> l >> r;
    long long dif = (r - l + n - 1) % n + 1;
    long long dif2 = (l - r + n - 1) % n + 1;
    s[l - 1].insert(dif2);
    s[l - 1 + n].insert(dif2);
    s[r - 1].insert(dif);
    s[r - 1 + n].insert(dif);
  }
  z_function();
  for (long long i = 1; i < n / 2; i++) {
    if (z[i] >= n / 2) {
      cout << "Yes" << endl;
      return 0;
    }
  }
  cout << "No" << endl;
  return 0;
}

### 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;
long long n, m;
long long z[200005];
set<long long> s[200005];
long long check(long long a, long long b) {
  if (s[a].size() == s[b].size()) {
    auto itr1 = s[a].begin(), itr2 = s[b].begin();
    while (itr1 != s[a].end()) {
      if ((*itr1) != (*itr2)) return 0;
      itr1++;
      itr2++;
    }
    return 1;
  }
  return 0;
}
long long z_function() {
  n = 2 * n;
  for (long long i = 1, l = 0, r = 0; i < n; ++i) {
    if (i <= r) z[i] = min(r - i + 1, z[i - l]);
    while (i + z[i] < n && check(z[i], i + z[i])) ++z[i];
    if (i + z[i] - 1 > r) l = i, r = i + z[i] - 1;
  }
  return 0;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> m;
  for (long long i = 0; i < m; i++) {
    long long l, r;
    cin >> l >> r;
    long long dif = (r - l + n - 1) % n + 1;
    long long dif2 = (l - r + n - 1) % n + 1;
    s[l - 1].insert(dif2);
    s[l - 1 + n].insert(dif2);
    s[r - 1].insert(dif);
    s[r - 1 + n].insert(dif);
  }
  z_function();
  for (long long i = 1; i < n / 2; i++) {
    if (z[i] >= n / 2) {
      cout << "Yes" << endl;
      return 0;
    }
  }
  cout << "No" << endl;
  return 0;
}

```