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 INF = 999999999999999999;
const double PI = acos(-1.0);
void stop() { exit(0); }
long long A[200005];
long long B[200005];
set<pair<long long, long long>> st;
int main() {
  ios::sync_with_stdio(false);
  long long n, m;
  cin >> n >> m;
  for (long long i = 0; i < m; i++) {
    cin >> A[i] >> B[i];
    st.insert(make_pair(A[i], B[i]));
    st.insert(make_pair(B[i], A[i]));
  }
  for (long long i = 1; i * i <= n; i++) {
    if (n % i == 0) {
      bool ye = 1;
      for (long long j = 0; j < m; j++) {
        long long d1 = A[j] + i;
        if (d1 > n) d1 -= n;
        long long d2 = B[j] + i;
        if (d2 > n) d2 -= n;
        if (st.find(make_pair(d1, d2)) == st.end() &&
            st.find(make_pair(d2, d1)) == st.end()) {
          ye = 0;
          break;
        }
      }
      if (ye) {
        cout << "Yes";
        stop();
      }
      long long d = n / i;
      if (d == n) continue;
      ye = 1;
      for (long long j = 0; j < m; j++) {
        long long d1 = A[j] + d;
        if (d1 > n) d1 -= n;
        long long d2 = B[j] + d;
        if (d2 > n) d2 -= n;
        if (st.find(make_pair(d1, d2)) == st.end() &&
            st.find(make_pair(d2, d1)) == st.end()) {
          ye = 0;
          break;
        }
      }
      if (ye) {
        cout << "Yes";
        stop();
      }
    }
  }
  cout << "No";
  stop();
}
