#include <bits/stdc++.h>
using namespace std;
vector<int> prefix_function(const vector<int>& s) {
  int n = s.size();
  vector<int> pi(n);
  for (int i = 1; i < n; ++i) {
    int j = pi[i - 1];
    while (j > 0 && s[i] != s[j]) j = pi[j - 1];
    if (s[i] == s[j]) ++j;
    pi[i] = j;
  }
  return pi;
}
int n;
set<int> f[100005];
int getLen(int a, int b) {
  int len = b - a;
  if (len <= 0) len += n;
  return len;
}
int go(int len) {
  if (f[len].size() == 0)
    return 1;
  else if (f[len].size() < 2)
    return -1;
  vector<int> diffs;
  int last = -1, fir = -1;
  for (int x : f[len]) {
    if (last == -1)
      fir = x;
    else
      diffs.push_back(getLen(last, x));
    last = x;
  }
  diffs.push_back(getLen(last, fir));
  vector<int> pref = prefix_function(diffs);
  int cand = diffs.size() - pref.back();
  if (diffs.size() % cand == 0 && cand < diffs.size()) {
    int sum = 0;
    for (int i = 0; i < cand; ++i) sum += diffs[i];
    return sum;
  }
  return -1;
}
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
int main() {
  ios::sync_with_stdio(false);
  int m;
  cin >> n >> m;
  for (int i = 0; i < m; ++i) {
    int a, b;
    cin >> a >> b;
    int len = getLen(a, b);
    if (2 * len > n || 2 * len == n && a > b) len = n - len, swap(a, b);
    f[len].insert(a - 1);
  }
  long long ans = 1;
  for (int i = 1; i < n; ++i) {
    if (2 * i == n) continue;
    int cur = go(i);
    if (cur == -1) {
      cout << "No" << endl;
      return 0;
    }
    ans = lcm(ans, cur);
    if (ans >= n) {
      cout << "No" << endl;
      return 0;
    }
  }
  if (n % 2 == 0) {
    int cur = go(n / 2);
    if (cur != -1 && lcm(ans, cur) < n || lcm(ans, n / 2) < n) {
    } else {
      cout << "No" << endl;
      return 0;
    }
  }
  cout << "Yes" << endl;
  return 0;
}

### Prompt

Create a solution in cpp for 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;
vector<int> prefix_function(const vector<int>& s) {
  int n = s.size();
  vector<int> pi(n);
  for (int i = 1; i < n; ++i) {
    int j = pi[i - 1];
    while (j > 0 && s[i] != s[j]) j = pi[j - 1];
    if (s[i] == s[j]) ++j;
    pi[i] = j;
  }
  return pi;
}
int n;
set<int> f[100005];
int getLen(int a, int b) {
  int len = b - a;
  if (len <= 0) len += n;
  return len;
}
int go(int len) {
  if (f[len].size() == 0)
    return 1;
  else if (f[len].size() < 2)
    return -1;
  vector<int> diffs;
  int last = -1, fir = -1;
  for (int x : f[len]) {
    if (last == -1)
      fir = x;
    else
      diffs.push_back(getLen(last, x));
    last = x;
  }
  diffs.push_back(getLen(last, fir));
  vector<int> pref = prefix_function(diffs);
  int cand = diffs.size() - pref.back();
  if (diffs.size() % cand == 0 && cand < diffs.size()) {
    int sum = 0;
    for (int i = 0; i < cand; ++i) sum += diffs[i];
    return sum;
  }
  return -1;
}
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
int main() {
  ios::sync_with_stdio(false);
  int m;
  cin >> n >> m;
  for (int i = 0; i < m; ++i) {
    int a, b;
    cin >> a >> b;
    int len = getLen(a, b);
    if (2 * len > n || 2 * len == n && a > b) len = n - len, swap(a, b);
    f[len].insert(a - 1);
  }
  long long ans = 1;
  for (int i = 1; i < n; ++i) {
    if (2 * i == n) continue;
    int cur = go(i);
    if (cur == -1) {
      cout << "No" << endl;
      return 0;
    }
    ans = lcm(ans, cur);
    if (ans >= n) {
      cout << "No" << endl;
      return 0;
    }
  }
  if (n % 2 == 0) {
    int cur = go(n / 2);
    if (cur != -1 && lcm(ans, cur) < n || lcm(ans, n / 2) < n) {
    } else {
      cout << "No" << endl;
      return 0;
    }
  }
  cout << "Yes" << endl;
  return 0;
}

```