#include <bits/stdc++.h>
using namespace std;
template <class t>
inline t read(t &x) {
  char c = getchar();
  bool f = 0;
  x = 0;
  while (!isdigit(c)) f |= c == '-', c = getchar();
  while (isdigit(c)) x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
  if (f) x = -x;
  return x;
}
template <class t, class... A>
inline void read(t &x, A &...a) {
  read(x);
  read(a...);
}
template <class t>
inline void write(t x) {
  if (x < 0)
    putchar('-'), write(-x);
  else {
    if (x > 9) write(x / 10);
    putchar('0' + x % 10);
  }
}
const int N = 2e5 + 5, mod = 1e9 + 7;
vector<int> g[N];
int n, h[N], m;
bool check(int d) {
  for (int i = 0; i < n; i++)
    if (h[i] != h[(i + d) % n]) return 0;
  return 1;
}
int calc(vector<int> &g) {
  int res = 0;
  for (int x : g) res = (res * 233ll + x) % mod;
  return res;
}
signed main() {
  read(n, m);
  for (int i = 1, x, y; i <= m; i++) {
    read(x, y);
    x--;
    y--;
    if (x > y) swap(x, y);
    g[x].push_back(y - x);
    g[y].push_back(n - y + x);
  }
  for (int i = 0; i < n; i++)
    sort((g[i]).begin(), (g[i]).end()), h[i] = calc(g[i]);
  for (int d = 1; d <= n / 2; d++)
    if (n % d == 0)
      if (check(d)) return puts("Yes");
  puts("No");
}

### Prompt

Please formulate 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;
template <class t>
inline t read(t &x) {
  char c = getchar();
  bool f = 0;
  x = 0;
  while (!isdigit(c)) f |= c == '-', c = getchar();
  while (isdigit(c)) x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
  if (f) x = -x;
  return x;
}
template <class t, class... A>
inline void read(t &x, A &...a) {
  read(x);
  read(a...);
}
template <class t>
inline void write(t x) {
  if (x < 0)
    putchar('-'), write(-x);
  else {
    if (x > 9) write(x / 10);
    putchar('0' + x % 10);
  }
}
const int N = 2e5 + 5, mod = 1e9 + 7;
vector<int> g[N];
int n, h[N], m;
bool check(int d) {
  for (int i = 0; i < n; i++)
    if (h[i] != h[(i + d) % n]) return 0;
  return 1;
}
int calc(vector<int> &g) {
  int res = 0;
  for (int x : g) res = (res * 233ll + x) % mod;
  return res;
}
signed main() {
  read(n, m);
  for (int i = 1, x, y; i <= m; i++) {
    read(x, y);
    x--;
    y--;
    if (x > y) swap(x, y);
    g[x].push_back(y - x);
    g[y].push_back(n - y + x);
  }
  for (int i = 0; i < n; i++)
    sort((g[i]).begin(), (g[i]).end()), h[i] = calc(g[i]);
  for (int d = 1; d <= n / 2; d++)
    if (n % d == 0)
      if (check(d)) return puts("Yes");
  puts("No");
}

```