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 size_t M = 1e9 + 7;
size_t getHash(const vector<int>& v) {
  size_t p = 1, result = 0;
  for (int x : v) {
    result += p * (x + 1);
    p *= 163841;
    result %= M;
    p %= M;
  }
  return result;
}
vector<int> z_function(vector<size_t> v) {
  int n = (int)v.size();
  vector<int> z(n);
  for (int i = 1, l = 0, r = 0; i < n; ++i) {
    if (i <= r) z[i] = min(r - i + 1, z[i - l]);
    while (i + z[i] < n && v[z[i]] == v[i + z[i]]) ++z[i];
    if (i + z[i] - 1 > r) l = i, r = i + z[i] - 1;
  }
  return z;
}
int main() {
  int n, m;
  cin >> n >> m;
  vector<pair<int, int>> h;
  for (int i = 0; i < m; i++) {
    int x, y;
    cin >> x >> y;
    h.push_back({x - 1, y > x ? y - x : n + y - x});
    h.push_back({y - 1, x > y ? x - y : n + x - y});
  }
  sort(h.begin(), h.end());
  vector<vector<int>> hh(n);
  for (auto& p : h) {
    hh[p.first].push_back(p.second);
  }
  vector<size_t> c(n * 2);
  for (int i = 0; i < 2 * n; i++) {
    if (i < n) {
      c[i] = getHash(hh[i]);
    } else {
      c[i] = c[i - n];
    }
  }
  auto z = z_function(c);
  bool ans = false;
  for (int i = 1; i < n; i++) {
    if (z[i] >= n) {
      ans = true;
    }
  }
  cout << (ans ? "Yes" : "No") << endl;
  return 0;
}
