#include <bits/stdc++.h>
using namespace std;
inline int read() {
  int x = 0;
  char c = getchar();
  while (c < '0' || c > '9') c = getchar();
  while (c >= '0' && c <= '9') x = (x << 3) + (x << 1) + c - '0', c = getchar();
  return x;
}
const int maxn = 1e5 + 10;
const int mod = 1e9 + 7;
const double eps = 1e-9;
int n, m;
int a[maxn], nt[maxn];
void Next() {
  nt[0] = -1;
  int j = 0, k = -1;
  while (j < n) {
    if (k == -1 || a[j] == a[k]) {
      ++j, ++k;
      nt[j] = k;
    } else
      k = nt[k];
  }
}
int main() {
  n = read(), m = read();
  int f = 0;
  for (int i = 0; i < m; ++i) {
    int u = read() - 1, v = read() - 1;
    if (u > v) swap(u, v);
    if (v - u <= n / 2)
      a[u]++, a[v]--;
    else
      a[0]++, a[u]--, a[v]++, a[n]--;
    if (v - u == n / 2) f++;
  }
  for (int i = 1; i < n; ++i) a[i] += a[i - 1];
  Next();
  int cycle = n - nt[n];
  if (cycle < n && n % cycle == 0)
    printf("Yes\n");
  else if (cycle == n && f == m && n % 2 == 0) {
    printf("Yes\n");
  } else
    printf("No\n");
}

### 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;
inline int read() {
  int x = 0;
  char c = getchar();
  while (c < '0' || c > '9') c = getchar();
  while (c >= '0' && c <= '9') x = (x << 3) + (x << 1) + c - '0', c = getchar();
  return x;
}
const int maxn = 1e5 + 10;
const int mod = 1e9 + 7;
const double eps = 1e-9;
int n, m;
int a[maxn], nt[maxn];
void Next() {
  nt[0] = -1;
  int j = 0, k = -1;
  while (j < n) {
    if (k == -1 || a[j] == a[k]) {
      ++j, ++k;
      nt[j] = k;
    } else
      k = nt[k];
  }
}
int main() {
  n = read(), m = read();
  int f = 0;
  for (int i = 0; i < m; ++i) {
    int u = read() - 1, v = read() - 1;
    if (u > v) swap(u, v);
    if (v - u <= n / 2)
      a[u]++, a[v]--;
    else
      a[0]++, a[u]--, a[v]++, a[n]--;
    if (v - u == n / 2) f++;
  }
  for (int i = 1; i < n; ++i) a[i] += a[i - 1];
  Next();
  int cycle = n - nt[n];
  if (cycle < n && n % cycle == 0)
    printf("Yes\n");
  else if (cycle == n && f == m && n % 2 == 0) {
    printf("Yes\n");
  } else
    printf("No\n");
}

```