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>
import java.io.*;
import java.util.*;
import java.util.stream.*;

public class Main {
  public static void main(String[] args) {
    boolean oj = System.getProperty("ONLINE_JUDGE") != null;
    try (BufferedReader reader = oj ? new BufferedReader( new InputStreamReader(System.in)) : new BufferedReader(new FileReader("input"));
         Writer writer = oj ? new OutputStreamWriter(System.out) : new FileWriter("output");
         PrintWriter out = new PrintWriter(writer);) {
      String line = null;
MAIN: while ((line = reader.readLine()) != null) {
        if (line.isEmpty()) {
          continue;
        }
        StringTokenizer tokenizer = new StringTokenizer(line);
        int n = Integer.parseInt(tokenizer.nextToken());
        int m = Integer.parseInt(tokenizer.nextToken());
        //debug(n + " " +  m);
        List<Integer> divisors = new ArrayList<>();
        divisors.add(1);
        for (int i = 2; i*i <= n; ++i) {
          if (n % i == 0) {
            divisors.add(i);
            divisors.add(n/i);
          }
        }
        divisors = divisors.stream().sorted().distinct().collect(Collectors.toList());
        Collections.reverse(divisors);

        int[] a = new int[m];
        int[] b = new int[m];
        List<Integer>[] vs = new List[n];
        for (int i = 0; i< n; ++i) {
          vs[i] = new ArrayList<>();
        }
        for (int i = 0; i< m; ++i) {
          line = reader.readLine();
          tokenizer = new StringTokenizer(line);
          a[i] = Integer.parseInt(tokenizer.nextToken()) -1 ;
          b[i] = Integer.parseInt(tokenizer.nextToken()) - 1;
          vs[a[i]].add(b[i]);
          vs[b[i]].add(a[i]);
        }
        for (int div : divisors) {
          boolean ok = true;
          //check div
          for (int i =0; i<m; ++i) {
            int anew = (a[i] + div) % n;
            int bnew = (b[i] + div) % n;
            if (!vs[anew].contains(bnew)) {
              ok = false;
              break;
            }
          }
          if (ok) {
            out.println("Yes");
            continue MAIN;
          }
        }
        out.println("No");
      }
    } catch (Exception e) {
      e.printStackTrace(System.err);
    }
  }

  static void debug(String s) {
    System.out.println(s);
  }
}
