#include <bits/stdc++.h>
using namespace std;
long long a[2001], d;
long double dp[1001][1001][2];
long long ans = 0;
long long min(long long a, long long b) {
  if (a < b)
    return a;
  else
    return b;
}
vector<long long> z_function(vector<vector<long long>> s) {
  long long n = (long long)s.size();
  vector<long long> z(n);
  z[0] = 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 && s[z[i]] == s[i + z[i]]) ++z[i];
    if (i + z[i] - 1 > r) l = i, r = i + z[i] - 1;
  }
  return z;
}
void fun() {
  long long i, j, k, x = 0, t, r, h, l, m, n, y;
  cin >> n >> m;
  vector<long long> d;
  vector<vector<long long>> v;
  v.assign(n, d);
  vector<long long> z;
  long long f;
  for (i = 0; i < m; i++) {
    cin >> l >> r;
    f = (r - l + n) % n;
    v[l - 1].push_back(f);
    v[r - 1].push_back(n - f);
    v[l - 1].push_back(n - f);
    v[r - 1].push_back(f);
  }
  for (i = 0; i < n; i++) {
    sort(v[i].begin(), v[i].end());
  }
  z = (z_function(v));
  vector<long long> c;
  c.push_back(1);
  for (long long i = 2; i <= sqrt(n); i++) {
    if (n % i == 0) {
      if (n / i == i)
        c.push_back(i);
      else {
        c.push_back(i);
        c.push_back(n / i);
      }
    }
  }
  for (i = 0; i < c.size(); i++) {
    for (j = 0; j < n; j += c[i]) {
      if (z[j] < c[i]) break;
    }
    if (j == n) {
      cout << "Yes" << endl;
      return;
    }
  }
  cout << "No" << endl;
}
int main() {
  long long t = 1;
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  while (t--) {
    fun();
  }
}

### Prompt

Your task is to create 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 a[2001], d;
long double dp[1001][1001][2];
long long ans = 0;
long long min(long long a, long long b) {
  if (a < b)
    return a;
  else
    return b;
}
vector<long long> z_function(vector<vector<long long>> s) {
  long long n = (long long)s.size();
  vector<long long> z(n);
  z[0] = 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 && s[z[i]] == s[i + z[i]]) ++z[i];
    if (i + z[i] - 1 > r) l = i, r = i + z[i] - 1;
  }
  return z;
}
void fun() {
  long long i, j, k, x = 0, t, r, h, l, m, n, y;
  cin >> n >> m;
  vector<long long> d;
  vector<vector<long long>> v;
  v.assign(n, d);
  vector<long long> z;
  long long f;
  for (i = 0; i < m; i++) {
    cin >> l >> r;
    f = (r - l + n) % n;
    v[l - 1].push_back(f);
    v[r - 1].push_back(n - f);
    v[l - 1].push_back(n - f);
    v[r - 1].push_back(f);
  }
  for (i = 0; i < n; i++) {
    sort(v[i].begin(), v[i].end());
  }
  z = (z_function(v));
  vector<long long> c;
  c.push_back(1);
  for (long long i = 2; i <= sqrt(n); i++) {
    if (n % i == 0) {
      if (n / i == i)
        c.push_back(i);
      else {
        c.push_back(i);
        c.push_back(n / i);
      }
    }
  }
  for (i = 0; i < c.size(); i++) {
    for (j = 0; j < n; j += c[i]) {
      if (z[j] < c[i]) break;
    }
    if (j == n) {
      cout << "Yes" << endl;
      return;
    }
  }
  cout << "No" << endl;
}
int main() {
  long long t = 1;
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  while (t--) {
    fun();
  }
}

```