#include <bits/stdc++.h>
using namespace std;
const int nn = 55;
int main() {
  int n, m, a, b;
  set<pair<int, int> > data;
  scanf("%d %d", &n, &m);
  for (int i = 0; i < m; i++) {
    scanf("%d %d", &a, &b);
    if (a > b) swap(a, b);
    data.insert(make_pair(a, b));
  }
  bool ans = 0;
  for (int i = 1; i <= sqrt(n); i++) {
    if (n % i == 0) {
      set<pair<int, int> > done;
      int k = i;
      bool cek = 1;
      for (set<pair<int, int> >::iterator j = data.begin(); j != data.end();
           ++j) {
        if (done.count(*j) == 0) {
          pair<int, int> tmp = *j;
          int x = tmp.first;
          int y = tmp.second;
          while (done.count(make_pair(x, y)) == 0 &&
                 data.count(make_pair(x, y)) == 1) {
            done.insert(make_pair(x, y));
            x += k;
            y += k;
            if (x > n) x -= n;
            if (y > n) y -= n;
            if (x > y) swap(x, y);
          }
          if (x != tmp.first || y != tmp.second) {
            cek = 0;
            break;
          }
        }
      }
      if (cek) ans = 1;
      done.clear();
      k = n / i;
      if (k == n) continue;
      cek = 1;
      for (set<pair<int, int> >::iterator j = data.begin(); j != data.end();
           ++j) {
        if (done.count(*j) == 0) {
          pair<int, int> tmp = *j;
          int x = tmp.first;
          int y = tmp.second;
          while (done.count(make_pair(x, y)) == 0 &&
                 data.count(make_pair(x, y)) == 1) {
            done.insert(make_pair(x, y));
            x += k;
            y += k;
            if (x > n) x -= n;
            if (y > n) y -= n;
            if (x > y) swap(x, y);
          }
          if (x != tmp.first || y != tmp.second) {
            cek = 0;
            break;
          }
        }
      }
      if (cek) ans = 1;
    }
  }
  if (ans)
    printf("Yes\n");
  else
    printf("No\n");
}

### Prompt

Create a solution in Cpp for 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;
const int nn = 55;
int main() {
  int n, m, a, b;
  set<pair<int, int> > data;
  scanf("%d %d", &n, &m);
  for (int i = 0; i < m; i++) {
    scanf("%d %d", &a, &b);
    if (a > b) swap(a, b);
    data.insert(make_pair(a, b));
  }
  bool ans = 0;
  for (int i = 1; i <= sqrt(n); i++) {
    if (n % i == 0) {
      set<pair<int, int> > done;
      int k = i;
      bool cek = 1;
      for (set<pair<int, int> >::iterator j = data.begin(); j != data.end();
           ++j) {
        if (done.count(*j) == 0) {
          pair<int, int> tmp = *j;
          int x = tmp.first;
          int y = tmp.second;
          while (done.count(make_pair(x, y)) == 0 &&
                 data.count(make_pair(x, y)) == 1) {
            done.insert(make_pair(x, y));
            x += k;
            y += k;
            if (x > n) x -= n;
            if (y > n) y -= n;
            if (x > y) swap(x, y);
          }
          if (x != tmp.first || y != tmp.second) {
            cek = 0;
            break;
          }
        }
      }
      if (cek) ans = 1;
      done.clear();
      k = n / i;
      if (k == n) continue;
      cek = 1;
      for (set<pair<int, int> >::iterator j = data.begin(); j != data.end();
           ++j) {
        if (done.count(*j) == 0) {
          pair<int, int> tmp = *j;
          int x = tmp.first;
          int y = tmp.second;
          while (done.count(make_pair(x, y)) == 0 &&
                 data.count(make_pair(x, y)) == 1) {
            done.insert(make_pair(x, y));
            x += k;
            y += k;
            if (x > n) x -= n;
            if (y > n) y -= n;
            if (x > y) swap(x, y);
          }
          if (x != tmp.first || y != tmp.second) {
            cek = 0;
            break;
          }
        }
      }
      if (cek) ans = 1;
    }
  }
  if (ans)
    printf("Yes\n");
  else
    printf("No\n");
}

```