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 long long N = 100005;
long long vn, en;
vector<long long> seq;
void init();
void process();
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  init();
  process();
  cout.flush();
  return 0;
}
void init() {
  cin >> vn >> en;
  vector<long long> G[N];
  for (long long i = 0; i < en; i++) {
    long long a, b;
    cin >> a >> b;
    a--, b--;
    G[a].push_back((b - a + vn) % vn);
    G[b].push_back((a - b + vn) % vn);
  }
  for (long long i = 0; i < vn; i++) {
    sort(G[i].begin(), G[i].end());
    seq.push_back(G[i].size());
    for (auto d : G[i]) seq.push_back(d);
  }
  seq.insert(seq.end(), seq.begin(), seq.end());
}
void process() {
  vector<long long> fail(seq.size());
  fail[0] = -1;
  for (long long i = 1, f = -1; i < fail.size(); i++) {
    while (f > -1 && seq[f + 1] != seq[i]) f = fail[f];
    if (seq[f + 1] == seq[i]) f++;
    fail[i] = f;
  }
  for (auto x : fail) {
    if (x == seq.size() / 2) {
      cout << "Yes" << '\n';
      return;
    }
  }
  cout << "No" << '\n';
}
