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>
#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const int MAXN = 1e5 + 5;
const char YES[] = "Yes";
const char NO[] = "No";
bool comp(const pair<int, int>& p1, const pair<int, int>& p2) {
  if (p1.first != p2.first) {
    return p1.first < p2.first;
  } else {
    return p1.second < p2.second;
  }
}
vector<int> getDiv(int n) {
  vector<int> ret;
  ret.push_back(1);
  for (int i = 2; i * i <= n; ++i) {
    if (n % i == 0) {
      ret.push_back(i);
      ret.push_back(n / i);
    }
  }
  return ret;
}
vector<pair<int, int> > ori;
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  vector<int> divs = getDiv(n);
  for (int i = 1; i <= m; ++i) {
    int a, b;
    scanf("%d%d", &a, &b);
    if (a > b) swap(a, b);
    ori.push_back({a, b});
  }
  sort(ori.begin(), ori.end());
  bool suc = false;
  for (auto f : divs) {
    vector<pair<int, int> > rot = ori;
    for (auto& p : rot) {
      p.first += f;
      if (p.first > n) {
        p.first -= n;
      }
      p.second += f;
      if (p.second > n) {
        p.second -= n;
      }
      if (p.first > p.second) swap(p.first, p.second);
    }
    sort(rot.begin(), rot.end());
    if (rot == ori) {
      suc = true;
      break;
    }
  }
  if (suc) {
    printf("%s\n", YES);
  } else {
    printf("%s\n", NO);
  }
}
