Problem: Vasya has an array a_1, a_2, ..., a_n.

You don't know this array, but he told you m facts about this array. The i-th fact is a triple of numbers t_i, l_i and r_i (0 ≤ t_i ≤ 1, 1 ≤ l_i < r_i ≤ n) and it means:

  * if t_i=1 then subbarray a_{l_i}, a_{l_i + 1}, ..., a_{r_i} is sorted in non-decreasing order; 
  * if t_i=0 then subbarray a_{l_i}, a_{l_i + 1}, ..., a_{r_i} is not sorted in non-decreasing order. A subarray is not sorted if there is at least one pair of consecutive elements in this subarray such that the former is greater than the latter. 



For example if a = [2, 1, 1, 3, 2] then he could give you three facts: t_1=1, l_1=2, r_1=4 (the subarray [a_2, a_3, a_4] = [1, 1, 3] is sorted), t_2=0, l_2=4, r_2=5 (the subarray [a_4, a_5] = [3, 2] is not sorted), and t_3=0, l_3=3, r_3=5 (the subarray [a_3, a_5] = [1, 3, 2] is not sorted).

You don't know the array a. Find any array which satisfies all the given facts.

Input

The first line contains two integers n and m (2 ≤ n ≤ 1000, 1 ≤ m ≤ 1000).

Each of the next m lines contains three integers t_i, l_i and r_i (0 ≤ t_i ≤ 1, 1 ≤ l_i < r_i ≤ n).

If t_i = 1 then subbarray a_{l_i}, a_{l_i + 1}, ... , a_{r_i} is sorted. Otherwise (if t_i = 0) subbarray a_{l_i}, a_{l_i + 1}, ... , a_{r_i} is not sorted.

Output

If there is no array that satisfies these facts in only line print NO (in any letter case).

If there is a solution, print YES (in any letter case). In second line print n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the array a, satisfying all the given facts. If there are multiple satisfying arrays you can print any of them.

Examples

Input


7 4
1 1 3
1 2 5
0 5 6
1 6 7


Output


YES
1 2 2 3 5 4 4


Input


4 2
1 1 4
0 2 3


Output


NO

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.*;


public class Cony {


	public static void main(String[] args) throws IOException
	{

		BufferedReader jk = new BufferedReader(new InputStreamReader(System.in)) ;
		PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out)) ; 
		StringTokenizer ana = new StringTokenizer(jk.readLine()) ; 

		int n = Integer.parseInt(ana.nextToken()) ; 
		int tab[]= new int[n] ;
		int m = Integer.parseInt(ana.nextToken()) ; 
		ArrayList<C> v0 = new ArrayList<>() ; 
		ArrayList<C> v1 = new ArrayList<>() ; 
		for(int i=0 ; i< m ;i++)
		{
			ana = new StringTokenizer(jk.readLine()) ; 
			int t = Integer.parseInt(ana.nextToken()) ; 
			int l = Integer.parseInt(ana.nextToken())-1 ; 
			int r= Integer.parseInt(ana.nextToken())-1 ; 
			if(t==0) 	v0.add(new C(l, r)) ;
			else 	v1.add(new C(l, r)) ;
		}
		if(v1.size()>0)
		{
			Collections.sort(v1);
			int l = v1.get(0).l ; int r = v1.get(0).r ;
			int debut = 2000 ; 
			for(int i=0 ;i< l ;i++)
			{
				tab[i]=debut ;debut-- ; 
			}

			for(int i=1 ; i<v1.size() ;i++)
			{
				int l1 = v1.get(i).l ;  int r1 = v1.get(i).r ;
				if(l1<=r) r=Integer.max(r1, r) ; 
				else
				{
					for(int j=l ; j<=r ; j++) tab[j]= debut ; 
					debut -- ; 
					for(int j= r+1 ; j<l1 ;j++)	
					{
						tab[j]= debut ;  debut -- ;
					}
					l= l1 ; r= r1 ; 
				}
			}
			for(int i= l ; i<=r ;i++) tab[i]=debut; 
			debut -- ; 
			for(int i=r+1 ; i<n ;i++)
			{
				tab[i]=debut;    debut -- ; 
			}
			boolean t = true;
			for(C ele : v0)
			{
				int l1= ele.l ;  int r1 = ele.r ;
				t=false ; 
				for(int i=l1 ; i<r1 ;i++)
				{
					if(tab[i]>tab[i+1])
					{
						t=true  ; break ; 
					}
				}
				if(t==false) break ;
			}
			if(!t) out.println("NO");
			else
			{
				out.println("YES");
				for(int i=0 ;i< n ;i++) out.print(tab[i]+" ");
			}
		}
		else
		{
			int eb= 2000 ; 
			out.println("YES");
			for(int i=0 ;i <n ;i++)
				{
				out.print(eb+" ");eb-- ; 
				}
		}
		
		
		out.flush();

	}
}
/*
 * 16 5
0 1 16
1 8 9
1 3 4
1 12 14
1 14 16*/

class C implements Comparable<C>
{
	int l , r ; 
	C(int l , int r )
	{
		this.l= l ; this.r= r; 
	}
	@Override
	public int compareTo(C o) {
		// TODO Auto-generated method stub
		if(this.l<o.l) return -1 ; 
		if(this.l>o.l) return 1 ; 
		return 0;
	}
}