#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;
int read() {
  int s = 0;
  char c = getchar(), lc = '+';
  while (c < '0' || '9' < c) lc = c, c = getchar();
  while ('0' <= c && c <= '9') s = s * 10 + c - '0', c = getchar();
  return lc == '-' ? -s : s;
}
void write(int x) {
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  if (x < 10)
    putchar(x + '0');
  else {
    write(x / 10);
    putchar(x % 10 + '0');
  }
}
void print(int x, char c = '\n') {
  write(x);
  putchar(c);
}
int siz[N];
vector<int> a[N];
bool check(int n, int k) {
  for (int i = 1; i <= n; i++)
    if (i * 2 != n)
      for (int j = 0; j < siz[i]; j++) {
        int pos =
            lower_bound(a[i].begin(), a[i].end(), a[i][j] + k) - a[i].begin();
        if (pos >= a[i].size() || a[i][pos] != a[i][j] + k) return 0;
      }
  return 1;
}
void ins(int l, int r, int n) {
  if (r < l) r += n;
  a[r - l].push_back(l);
}
signed main() {
  int n = read(), m = read();
  for (int i = 1; i <= m; i++) {
    int l = read(), r = read();
    ins(l, r, n);
    ins(r, l, n);
  }
  for (int i = 1; i <= n; i++) sort(a[i].begin(), a[i].end());
  for (int i = 1; i <= n; i++) {
    siz[i] = a[i].size();
    for (int j = 0; j < siz[i]; j++) a[i].push_back(a[i][j] + n);
    if (i * 2 == n)
      for (int j = 0; j < siz[i]; j++) a[i].push_back(a[i][j] + i);
  }
  for (int i = 1; i <= n; i++) sort(a[i].begin(), a[i].end());
  int nn = n;
  for (int k = 2; k <= sqrt(nn); k++)
    if (nn % k == 0) {
      while (nn % k == 0) nn /= k;
      if (check(n, n / k)) return puts("Yes"), 0;
    }
  if (nn > 1)
    if (check(n, n / nn)) return puts("Yes"), 0;
  puts("No");
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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;
const int N = 2e5 + 5;
int read() {
  int s = 0;
  char c = getchar(), lc = '+';
  while (c < '0' || '9' < c) lc = c, c = getchar();
  while ('0' <= c && c <= '9') s = s * 10 + c - '0', c = getchar();
  return lc == '-' ? -s : s;
}
void write(int x) {
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  if (x < 10)
    putchar(x + '0');
  else {
    write(x / 10);
    putchar(x % 10 + '0');
  }
}
void print(int x, char c = '\n') {
  write(x);
  putchar(c);
}
int siz[N];
vector<int> a[N];
bool check(int n, int k) {
  for (int i = 1; i <= n; i++)
    if (i * 2 != n)
      for (int j = 0; j < siz[i]; j++) {
        int pos =
            lower_bound(a[i].begin(), a[i].end(), a[i][j] + k) - a[i].begin();
        if (pos >= a[i].size() || a[i][pos] != a[i][j] + k) return 0;
      }
  return 1;
}
void ins(int l, int r, int n) {
  if (r < l) r += n;
  a[r - l].push_back(l);
}
signed main() {
  int n = read(), m = read();
  for (int i = 1; i <= m; i++) {
    int l = read(), r = read();
    ins(l, r, n);
    ins(r, l, n);
  }
  for (int i = 1; i <= n; i++) sort(a[i].begin(), a[i].end());
  for (int i = 1; i <= n; i++) {
    siz[i] = a[i].size();
    for (int j = 0; j < siz[i]; j++) a[i].push_back(a[i][j] + n);
    if (i * 2 == n)
      for (int j = 0; j < siz[i]; j++) a[i].push_back(a[i][j] + i);
  }
  for (int i = 1; i <= n; i++) sort(a[i].begin(), a[i].end());
  int nn = n;
  for (int k = 2; k <= sqrt(nn); k++)
    if (nn % k == 0) {
      while (nn % k == 0) nn /= k;
      if (check(n, n / k)) return puts("Yes"), 0;
    }
  if (nn > 1)
    if (check(n, n / nn)) return puts("Yes"), 0;
  puts("No");
  return 0;
}

```