#include <bits/stdc++.h>
using namespace std;
int n, m;
vector<pair<int, int> > a[200010];
bool F;
pair<int, int> b[200010];
int c1;
bool work(int x) {
  if (x == n) return 0;
  for (int l = 1; l <= n; ++l) {
    c1 = 0;
    for (auto it : a[l]) b[++c1] = make_pair(it.first + x, it.second + x);
    for (int i = 1; i <= c1; ++i) {
      if (b[i].first > n) b[i].first -= n;
      if (b[i].second > n) b[i].second -= n;
      if (b[i].first > b[i].second) swap(b[i].first, b[i].second);
    }
    sort(b + 1, b + c1 + 1);
    for (int i = 1; i <= c1; ++i)
      if (b[i] != a[l][i - 1]) return 0;
  }
  return 1;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; ++i) {
    int x, y, z;
    scanf("%d%d", &x, &y);
    if (x > y) swap(x, y);
    z = min(y - x, n - y + x);
    a[z].push_back(make_pair(x, y));
  }
  for (int i = 1; i <= n; ++i) sort(a[i].begin(), a[i].end());
  for (int i = 1; i * i <= n; ++i)
    if (n % i == 0) F |= work(i), F |= work(n / i);
  puts(F ? "YES" : "NO");
  return 0;
}

### Prompt

Generate 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;
int n, m;
vector<pair<int, int> > a[200010];
bool F;
pair<int, int> b[200010];
int c1;
bool work(int x) {
  if (x == n) return 0;
  for (int l = 1; l <= n; ++l) {
    c1 = 0;
    for (auto it : a[l]) b[++c1] = make_pair(it.first + x, it.second + x);
    for (int i = 1; i <= c1; ++i) {
      if (b[i].first > n) b[i].first -= n;
      if (b[i].second > n) b[i].second -= n;
      if (b[i].first > b[i].second) swap(b[i].first, b[i].second);
    }
    sort(b + 1, b + c1 + 1);
    for (int i = 1; i <= c1; ++i)
      if (b[i] != a[l][i - 1]) return 0;
  }
  return 1;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; ++i) {
    int x, y, z;
    scanf("%d%d", &x, &y);
    if (x > y) swap(x, y);
    z = min(y - x, n - y + x);
    a[z].push_back(make_pair(x, y));
  }
  for (int i = 1; i <= n; ++i) sort(a[i].begin(), a[i].end());
  for (int i = 1; i * i <= n; ++i)
    if (n % i == 0) F |= work(i), F |= work(n / i);
  puts(F ? "YES" : "NO");
  return 0;
}

```