#include <bits/stdc++.h>
using namespace std;
int n, m, w;
vector<vector<int> > seg;
int str[1000005];
int Z[1000005];
void getZarr() {
  int n = 2 * w;
  int L, R, k;
  L = R = 0;
  for (int i = 1; i < n; ++i) {
    if (i > R) {
      L = R = i;
      while (R < n && str[R - L] == str[R]) R++;
      Z[i] = R - L;
      R--;
    } else {
      k = i - L;
      if (Z[k] < R - i + 1)
        Z[i] = Z[k];
      else {
        L = i;
        while (R < n && str[R - L] == str[R]) R++;
        Z[i] = R - L;
        R--;
      }
    }
  }
}
int main(void) {
  cin >> n >> m;
  seg.resize(n + 1);
  for (int i = 0; i < m; i++) {
    int u, v;
    cin >> u >> v;
    if (u > v) {
      int tmp = u;
      u = v;
      v = tmp;
    }
    seg[u].push_back(v - u);
    seg[v].push_back(u + n - v);
  }
  int ptr = 0;
  for (int i = 1; i <= n; i++) {
    if (seg[i].size() == 0) {
      str[ptr++] = 0;
      continue;
    }
    sort(seg[i].begin(), seg[i].end());
    for (int j = 0; j < seg[i].size(); j++) str[ptr++] = seg[i][j];
  }
  w = ptr;
  for (int i = 0; i < w; i++) str[i + w] = str[i];
  getZarr();
  for (int i = 1; i < w; i++) {
    if (Z[i] >= w) {
      cout << "Yes" << endl;
      return 0;
    }
  }
  cout << "No" << endl;
}

### 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;
int n, m, w;
vector<vector<int> > seg;
int str[1000005];
int Z[1000005];
void getZarr() {
  int n = 2 * w;
  int L, R, k;
  L = R = 0;
  for (int i = 1; i < n; ++i) {
    if (i > R) {
      L = R = i;
      while (R < n && str[R - L] == str[R]) R++;
      Z[i] = R - L;
      R--;
    } else {
      k = i - L;
      if (Z[k] < R - i + 1)
        Z[i] = Z[k];
      else {
        L = i;
        while (R < n && str[R - L] == str[R]) R++;
        Z[i] = R - L;
        R--;
      }
    }
  }
}
int main(void) {
  cin >> n >> m;
  seg.resize(n + 1);
  for (int i = 0; i < m; i++) {
    int u, v;
    cin >> u >> v;
    if (u > v) {
      int tmp = u;
      u = v;
      v = tmp;
    }
    seg[u].push_back(v - u);
    seg[v].push_back(u + n - v);
  }
  int ptr = 0;
  for (int i = 1; i <= n; i++) {
    if (seg[i].size() == 0) {
      str[ptr++] = 0;
      continue;
    }
    sort(seg[i].begin(), seg[i].end());
    for (int j = 0; j < seg[i].size(); j++) str[ptr++] = seg[i][j];
  }
  w = ptr;
  for (int i = 0; i < w; i++) str[i + w] = str[i];
  getZarr();
  for (int i = 1; i < w; i++) {
    if (Z[i] >= w) {
      cout << "Yes" << endl;
      return 0;
    }
  }
  cout << "No" << endl;
}

```