#include <bits/stdc++.h>
using namespace std;
unsigned long long min(unsigned long long x, unsigned long long y) {
  if (x < y) return x;
  return y;
}
unsigned long long max(unsigned long long x, unsigned long long y) {
  if (x < y) return y;
  return x;
}
long long min(long long x, long long y) {
  if (x < y) return x;
  return y;
}
long long max(long long x, long long y) {
  if (x < y) return y;
  return x;
}
double min(double x, double y) {
  if (x < y) return x;
  return y;
}
double max(double x, double y) {
  if (x < y) return y;
  return x;
}
unsigned long long gcd(unsigned long long x, unsigned long long y) {
  if (!x) return y;
  if (!y) return x;
  if (x > y) swap(x, y);
  return gcd(x, y % x);
}
unsigned long long inv(unsigned long long a, unsigned long long c) {
  if (a == 1) {
    return 1;
  }
  return ((c - (c / a)) * inv(c % a, c)) % c;
}
int n, m, a, b, lps[100005] = {0};
vector<vector<int> > str, sub(100005);
void fill() {
  int i = 1, j = 0;
  while (i < n) {
    if (sub[j] == sub[i]) {
      lps[i] = j + 1;
      i++;
      j++;
    } else {
      if (!j) {
        i++;
      } else {
        j = lps[j - 1];
      }
    }
  }
}
int kmp() {
  int N = str.size();
  int i = 0, j = 0;
  while ((i < N) && (j < n)) {
    if (str[i] == sub[j]) {
      i++;
      j++;
    } else {
      if (!j)
        i++;
      else
        j = lps[j - 1];
    }
  }
  if (j < n) return 0;
  return 1;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    cin >> a >> b;
    if (a > b) swap(a, b);
    sub[a - 1].push_back(b - a);
    sub[b - 1].push_back(a - b + n);
  }
  for (int i = 0; i < n; i++) {
    sort(sub[i].begin(), sub[i].end());
  }
  for (int i = 1; i < n; i++) {
    str.push_back(sub[i]);
  }
  for (int i = 0; i < n - 1; i++) {
    str.push_back(sub[i]);
  }
  fill();
  if (kmp()) {
    cout << "Yes\n";
  } else {
    cout << "No\n";
  }
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
unsigned long long min(unsigned long long x, unsigned long long y) {
  if (x < y) return x;
  return y;
}
unsigned long long max(unsigned long long x, unsigned long long y) {
  if (x < y) return y;
  return x;
}
long long min(long long x, long long y) {
  if (x < y) return x;
  return y;
}
long long max(long long x, long long y) {
  if (x < y) return y;
  return x;
}
double min(double x, double y) {
  if (x < y) return x;
  return y;
}
double max(double x, double y) {
  if (x < y) return y;
  return x;
}
unsigned long long gcd(unsigned long long x, unsigned long long y) {
  if (!x) return y;
  if (!y) return x;
  if (x > y) swap(x, y);
  return gcd(x, y % x);
}
unsigned long long inv(unsigned long long a, unsigned long long c) {
  if (a == 1) {
    return 1;
  }
  return ((c - (c / a)) * inv(c % a, c)) % c;
}
int n, m, a, b, lps[100005] = {0};
vector<vector<int> > str, sub(100005);
void fill() {
  int i = 1, j = 0;
  while (i < n) {
    if (sub[j] == sub[i]) {
      lps[i] = j + 1;
      i++;
      j++;
    } else {
      if (!j) {
        i++;
      } else {
        j = lps[j - 1];
      }
    }
  }
}
int kmp() {
  int N = str.size();
  int i = 0, j = 0;
  while ((i < N) && (j < n)) {
    if (str[i] == sub[j]) {
      i++;
      j++;
    } else {
      if (!j)
        i++;
      else
        j = lps[j - 1];
    }
  }
  if (j < n) return 0;
  return 1;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    cin >> a >> b;
    if (a > b) swap(a, b);
    sub[a - 1].push_back(b - a);
    sub[b - 1].push_back(a - b + n);
  }
  for (int i = 0; i < n; i++) {
    sort(sub[i].begin(), sub[i].end());
  }
  for (int i = 1; i < n; i++) {
    str.push_back(sub[i]);
  }
  for (int i = 0; i < n - 1; i++) {
    str.push_back(sub[i]);
  }
  fill();
  if (kmp()) {
    cout << "Yes\n";
  } else {
    cout << "No\n";
  }
  return 0;
}

```