#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5;
struct Node {
  int x, y;
};
vector<Node> hh[N];
bool cmp(Node a, Node b) { return a.x < b.x; }
int smg[N];
map<pair<int, int>, int> flag;
int n, m;
bool solve(int x) {
  for (int i = 1; i <= n / 2; i++) {
    if (hh[i].size()) {
      for (int j = 0; j < hh[i].size(); j++) {
        int tep = hh[i][j].x + x;
        while (tep > n) tep -= n;
        if (flag[make_pair(tep, i)])
          continue;
        else {
          return 0;
        }
      }
    }
  }
  return 1;
}
vector<int> ans;
int main() {
  scanf("%d%d", &n, &m);
  bool f = 0;
  Node tep;
  for (int i = 1; i <= m; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    if (y > x) swap(x, y);
    if (x - y > y + n - x) {
      tep.x = x;
      tep.y = y;
      hh[y + n - x].push_back(tep);
      flag[make_pair(x, y + n - x)] = 1;
      if ((n % 2) == 0 && y + n - x == n / 2) flag[make_pair(y, y + n - x)] = 1;
    } else {
      tep.x = y;
      tep.y = x;
      hh[x - y].push_back(tep);
      flag[make_pair(y, x - y)] = 1;
      if ((n % 2) == 0 && x - y == n / 2) flag[make_pair(x, x - y)] = 1;
    }
  }
  for (int i = 1; i <= n / 2; i++) {
    if ((n % i) == 0) ans.push_back(i);
  }
  for (int i = 0; i < ans.size(); i++) {
    if (solve(ans[i])) {
      f = 1;
      printf("Yes\n");
      break;
    }
  }
  if (f == 0) printf("No\n");
  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 = 1e5 + 5;
struct Node {
  int x, y;
};
vector<Node> hh[N];
bool cmp(Node a, Node b) { return a.x < b.x; }
int smg[N];
map<pair<int, int>, int> flag;
int n, m;
bool solve(int x) {
  for (int i = 1; i <= n / 2; i++) {
    if (hh[i].size()) {
      for (int j = 0; j < hh[i].size(); j++) {
        int tep = hh[i][j].x + x;
        while (tep > n) tep -= n;
        if (flag[make_pair(tep, i)])
          continue;
        else {
          return 0;
        }
      }
    }
  }
  return 1;
}
vector<int> ans;
int main() {
  scanf("%d%d", &n, &m);
  bool f = 0;
  Node tep;
  for (int i = 1; i <= m; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    if (y > x) swap(x, y);
    if (x - y > y + n - x) {
      tep.x = x;
      tep.y = y;
      hh[y + n - x].push_back(tep);
      flag[make_pair(x, y + n - x)] = 1;
      if ((n % 2) == 0 && y + n - x == n / 2) flag[make_pair(y, y + n - x)] = 1;
    } else {
      tep.x = y;
      tep.y = x;
      hh[x - y].push_back(tep);
      flag[make_pair(y, x - y)] = 1;
      if ((n % 2) == 0 && x - y == n / 2) flag[make_pair(x, x - y)] = 1;
    }
  }
  for (int i = 1; i <= n / 2; i++) {
    if ((n % i) == 0) ans.push_back(i);
  }
  for (int i = 0; i < ans.size(); i++) {
    if (solve(ans[i])) {
      f = 1;
      printf("Yes\n");
      break;
    }
  }
  if (f == 0) printf("No\n");
  return 0;
}

```