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.util.*;
import java.io.*;

public class ChaldniFigure {
	static BufferedReader br;
	static StringTokenizer tokenizer;

	public static void main(String[] args) throws Exception {
		br = new BufferedReader(new InputStreamReader(System.in));
		int n = nextInt();
		int k = nextInt();
		long[] arr = new long[n];
		HashSet<Integer>[] arr1 = new HashSet[n];
		for(int i = 0; i < arr1.length; i++)
			arr1[i] = new HashSet<Integer>();
		for(int i = 0; i < k; i++) {
			int t = nextInt();
			int t1 = nextInt();
			if(t > t1) {
				int t2 = t;
				t = t1;
				t1 = t2;
			}
			arr[t - 1]++;
			arr[t1 - 1]++;
			arr1[t - 1].add(t1 - 1);
		}
		long[] pSum = new long[n];
		long[] pNum = new long[n];
		pNum[0] = arr[0];
		pSum[0] = arr[0];
		for(int i = 1; i < n; i++)
			pSum[i] = pSum[i - 1] + (arr[i] * (i + 1));
		for(int i = 1; i < n; i++)
			pNum[i] = pNum[i - 1] + arr[i];
		long curr = pSum[n - 1];
		outer:
		for(int i = 1; i < n; i++) {
			long t = pSum[n - 1];
			int l = n - i + 1;
			t -= (pNum[n - 1] - pNum[l - 2]) * n;
			t += pNum[n - 1] * i;
			if(t == curr) {
				for(int j = 0; j < n; j++) {
					Iterator<Integer> it = arr1[j].iterator();
					while(it.hasNext()) {
						int t1 = it.next();
						if(!arr1[Math.min((t1 + i) % n, (j + i) % n)].contains(Math.max((t1 + i) % n, (j + i) % n))) {
							continue outer;
						
						}
					}
				}
				System.out.println("Yes");
				return;
			}
		}
		System.out.println("No");
	}

	public static String next() throws IOException {
		while (tokenizer == null || !tokenizer.hasMoreTokens()) {
			String line = br.readLine();
			if (line == null)
				throw new IOException();
			tokenizer = new StringTokenizer(line);
		}
		return tokenizer.nextToken();
	}

	public static int nextInt() throws IOException {
		return Integer.parseInt(next());
	}

	public static long nextLong() throws IOException {
		return Long.parseLong(next());
	}

	public static double nextDouble() throws IOException {
		return Double.parseDouble(next());
	}
}
