#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 5, mod = 998244353;
int n, m;
set<pair<int, int> > mds;
int check(int x, int ff) {
  int nxt;
  if (ff != 1) {
    for (set<pair<int, int> >::iterator it = mds.begin(); it != mds.end();
         it++) {
      nxt = (it->first + x) % (2 * n);
      if (mds.find(make_pair(nxt, it->second)) == mds.end()) return 0;
    }
    return 1;
  } else {
    for (set<pair<int, int> >::iterator it = mds.begin(); it != mds.end();
         it++) {
      nxt = (it->first + x) % (2 * n);
      if (mds.find(make_pair(nxt, it->second)) == mds.end() &&
          mds.find(make_pair(nxt % n, it->second)) == mds.end())
        return 0;
    }
    return 1;
  }
}
vector<int> cnt[400000 + 50];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  int u, v, mid, dif, mdf = 0;
  for (int i = 1; i <= m; i++) {
    cin >> u >> v;
    u--, v--;
    if (abs(u - v) > n - (abs(u - v))) {
      dif = n - (abs(u - v));
      mid = (u + v + n);
      if (mid >= 2 * n) mid -= 2 * n;
    } else {
      dif = abs(u - v);
      mid = (u + v);
    }
    mds.insert(make_pair(mid, dif));
    cnt[dif].push_back(mid);
    mdf = max(mdf, dif);
  }
  int flag = 0;
  for (int i = 1; i <= mdf; i++) {
    if (cnt[i].size() != 0) {
      if (2 * i == n) {
        flag = check(n, 1);
      } else {
        int cc = cnt[i].size();
        for (int j = 1; j * j <= cc; j++) {
          if (cc % j == 0) {
            if (cc != j * j) {
              if (n % j == 0 && j != 1) flag = max(flag, check(2 * (n / j), 0));
              if (n % (cc / j) == 0)
                flag = max(flag, check(2 * (n / (cc / j)), 0));
            } else {
              flag = max(flag, check(2 * j, 0));
            }
          }
        }
      }
      break;
    }
  }
  if (flag) {
    cout << "Yes";
  } else {
    cout << "No";
  }
  return 0;
}

### Prompt

In CPP, your task is to solve 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 maxn = 1e5 + 5, mod = 998244353;
int n, m;
set<pair<int, int> > mds;
int check(int x, int ff) {
  int nxt;
  if (ff != 1) {
    for (set<pair<int, int> >::iterator it = mds.begin(); it != mds.end();
         it++) {
      nxt = (it->first + x) % (2 * n);
      if (mds.find(make_pair(nxt, it->second)) == mds.end()) return 0;
    }
    return 1;
  } else {
    for (set<pair<int, int> >::iterator it = mds.begin(); it != mds.end();
         it++) {
      nxt = (it->first + x) % (2 * n);
      if (mds.find(make_pair(nxt, it->second)) == mds.end() &&
          mds.find(make_pair(nxt % n, it->second)) == mds.end())
        return 0;
    }
    return 1;
  }
}
vector<int> cnt[400000 + 50];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  int u, v, mid, dif, mdf = 0;
  for (int i = 1; i <= m; i++) {
    cin >> u >> v;
    u--, v--;
    if (abs(u - v) > n - (abs(u - v))) {
      dif = n - (abs(u - v));
      mid = (u + v + n);
      if (mid >= 2 * n) mid -= 2 * n;
    } else {
      dif = abs(u - v);
      mid = (u + v);
    }
    mds.insert(make_pair(mid, dif));
    cnt[dif].push_back(mid);
    mdf = max(mdf, dif);
  }
  int flag = 0;
  for (int i = 1; i <= mdf; i++) {
    if (cnt[i].size() != 0) {
      if (2 * i == n) {
        flag = check(n, 1);
      } else {
        int cc = cnt[i].size();
        for (int j = 1; j * j <= cc; j++) {
          if (cc % j == 0) {
            if (cc != j * j) {
              if (n % j == 0 && j != 1) flag = max(flag, check(2 * (n / j), 0));
              if (n % (cc / j) == 0)
                flag = max(flag, check(2 * (n / (cc / j)), 0));
            } else {
              flag = max(flag, check(2 * j, 0));
            }
          }
        }
      }
      break;
    }
  }
  if (flag) {
    cout << "Yes";
  } else {
    cout << "No";
  }
  return 0;
}

```