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.*;


public class May {
 
	
	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int m = sc.nextInt();
		Set<Pair> set = new HashSet<>();
		
		for(int i=0;i<m;++i) {
			int x = sc.nextInt(); --x;
			int y = sc.nextInt(); --y;
			set.add(new Pair(x,y));
			
		}
		List<Integer> list = div(n);
		
		if(solve(set,list,n)) System.out.println("YES");
		else System.out.println("NO");
	}
	
	private static boolean solve(Set<Pair> set, List<Integer> list,int n) {
		for(int x : list) {
			boolean yes = true;
		for(Pair p : set) {
			  int nextKey = (p.x+x)%n;
			  int nextVal = (p.y+x)%n;
			  
			  if((set.contains(new Pair(nextKey,nextVal))) || (set.contains(new Pair(nextVal,nextKey)))){
				  continue;
			  }else {
				  yes = false;
				  break;
			  }
			
		   }
		if(yes) {
			//System.out.println(x);
			return true;
		}
		}
		return false;
	}

	static List<Integer> div(int n){
		List<Integer> list = new ArrayList<>();
		list.add(1);
		for(int i=2;i*i<=n;++i) {
			if(n%i == 0) {
				if(i == n/i) {
					list.add(i);
				}else {
					list.add(i);
					list.add(n/i);
				}
			}
		}
		return list;
	}
	
	static class Pair{
    	int x;
    	int y;
    	Pair(int x,int y){
    		this.x=x;
    		this.y=y;
    	}
    	@Override
        public int hashCode() {
            final int prime = 31;
           int  result = prime * x + y; 
            return result;
        }
    	
    	 @Override
    	    public boolean equals(Object obj) {
    	        if (this == obj)
    	            return true;
    	        if (obj == null)
    	            return false;
    	        if (getClass() != obj.getClass())
    	            return false;
    	        Pair other = (Pair) obj;
    	        if (x != other.x && y != other.y)
    	            return false;
    	        return true;
    	    }
    	
    }
}
