#include <bits/stdc++.h>
using std::map;
using std::max;
using std::vector;
const int MAXN = 1e5 + 20, MAXM = 2e5 + 20;
int read() {
  int x = 0;
  char c = getchar();
  bool f = 0;
  while (c < '0' || c > '9') (c == '-') ? f = 1, c = getchar() : c = getchar();
  while (c >= '0' && c <= '9')
    x = (x << 1) + (x << 3) + (48 ^ c), c = getchar();
  return (f) ? -x : x;
}
vector<int> arr[MAXN / 2];
vector<int> str[MAXN];
int ID;
map<int, int> son[MAXM];
int vis[MAXM], tot;
int insert(const int &x, const int &cur, vector<int> &Str) {
  if (cur == (int)Str.size() - 1) {
    if (!vis[x])
      return vis[x] = ++ID;
    else
      return vis[x];
  } else {
    if (!son[x].count(Str[cur + 1]))
      son[x].insert(std::pair<int, int>(Str[cur + 1], ++tot));
    return insert(son[x].find(Str[cur + 1])->second, cur + 1, Str);
  }
}
inline int dist(const int &a, const int &b, const int &n) {
  if (a < b)
    return b - a;
  else
    return b + (n - a);
}
int s[MAXN], pi[MAXN];
int main() {
  int n = read(), m = read();
  for (int i = 0; i < m; ++i) {
    int a = read(), b = read();
    int Dist = dist(a, b, n);
    if (Dist > n / 2) a ^= b ^= a ^= b, Dist = dist(a, b, n);
    arr[Dist].push_back(a);
    if (Dist == n / 2 && n % 2 == 0) arr[Dist].push_back(b);
  }
  for (int i = 1; i <= n; ++i) str[i].push_back(-1);
  for (int i = 0; i <= n / 2; ++i)
    for (int j : arr[i]) str[j].push_back(i);
  for (int i = 1; i <= n; ++i) s[i] = insert(0, -1, str[i]);
  int mx = 1;
  for (int i = 2; i <= n; ++i) {
    int j = pi[i - 1];
    while (j && s[j + 1] != s[i]) j = pi[j];
    if (s[j + 1] == s[i])
      pi[i] = j + 1;
    else
      pi[i] = 0;
  }
  for (int cur = pi[n]; cur; cur = pi[cur]) {
    int cur_len = n - cur;
    if (n % cur_len == 0) mx = max(mx, n / cur_len);
  }
  if (mx > 1)
    puts("Yes");
  else
    puts("No");
}

### Prompt

Your challenge is to write 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 std::map;
using std::max;
using std::vector;
const int MAXN = 1e5 + 20, MAXM = 2e5 + 20;
int read() {
  int x = 0;
  char c = getchar();
  bool f = 0;
  while (c < '0' || c > '9') (c == '-') ? f = 1, c = getchar() : c = getchar();
  while (c >= '0' && c <= '9')
    x = (x << 1) + (x << 3) + (48 ^ c), c = getchar();
  return (f) ? -x : x;
}
vector<int> arr[MAXN / 2];
vector<int> str[MAXN];
int ID;
map<int, int> son[MAXM];
int vis[MAXM], tot;
int insert(const int &x, const int &cur, vector<int> &Str) {
  if (cur == (int)Str.size() - 1) {
    if (!vis[x])
      return vis[x] = ++ID;
    else
      return vis[x];
  } else {
    if (!son[x].count(Str[cur + 1]))
      son[x].insert(std::pair<int, int>(Str[cur + 1], ++tot));
    return insert(son[x].find(Str[cur + 1])->second, cur + 1, Str);
  }
}
inline int dist(const int &a, const int &b, const int &n) {
  if (a < b)
    return b - a;
  else
    return b + (n - a);
}
int s[MAXN], pi[MAXN];
int main() {
  int n = read(), m = read();
  for (int i = 0; i < m; ++i) {
    int a = read(), b = read();
    int Dist = dist(a, b, n);
    if (Dist > n / 2) a ^= b ^= a ^= b, Dist = dist(a, b, n);
    arr[Dist].push_back(a);
    if (Dist == n / 2 && n % 2 == 0) arr[Dist].push_back(b);
  }
  for (int i = 1; i <= n; ++i) str[i].push_back(-1);
  for (int i = 0; i <= n / 2; ++i)
    for (int j : arr[i]) str[j].push_back(i);
  for (int i = 1; i <= n; ++i) s[i] = insert(0, -1, str[i]);
  int mx = 1;
  for (int i = 2; i <= n; ++i) {
    int j = pi[i - 1];
    while (j && s[j + 1] != s[i]) j = pi[j];
    if (s[j + 1] == s[i])
      pi[i] = j + 1;
    else
      pi[i] = 0;
  }
  for (int cur = pi[n]; cur; cur = pi[cur]) {
    int cur_len = n - cur;
    if (n % cur_len == 0) mx = max(mx, n / cur_len);
  }
  if (mx > 1)
    puts("Yes");
  else
    puts("No");
}

```