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.util.*;
import java.io.*;
 
public class Solution{
 
	public static Integer INT(String s){
		return Integer.parseInt(s);
	}
 
	public static Long LONG(String s){
		return Long.parseLong(s);
	}
 
	//====================================================================================================================
 
 
	static class Pair implements Comparable<Pair>{
		int t, l, r;
		Pair(int t, int l, int r){
			this.t=t;
			this.l=l;
			this.r=r;
		}

		public int compareTo(Pair obj){
			if(this.l==obj.l)
				return this.r-obj.r;
			return this.l-obj.l;
		}

		public String toString(){
			return "("+t+", "+l+", "+r+")";
		}
	}
 
 	public static boolean check(Pair node, ArrayList<Pair> fact1){

 		for(Pair item: fact1)
 			if((item.l<=node.l && node.l<=item.r) && (item.l<=node.r && node.r<=item.r))
 				return false;
 		return true;
 	}

 	public static ArrayList<Pair> compress(ArrayList<Pair> fact1){


 		ArrayList<Pair> new_fact1=new ArrayList<>();

 		if(fact1.size()==0)
 			return new_fact1;

 		Collections.sort(fact1);

 		int start=-1, end=-1;
 		for(Pair item: fact1){
 			if(end==-1){
 				start=item.l;
 				end=item.r;
 			}
 			if(item.l<=end)
 				end=Math.max(end, item.r);
 			else{
 				new_fact1.add(new Pair(1, start, end));
 				start=item.l;
 				end=item.r;
 			}
 		}
 		new_fact1.add(new Pair(1, start, end));
 		return new_fact1;
 	}
 
	public static void main(String args[])throws IOException{
 
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));	Scanner in=new Scanner(System.in);	StringBuilder out=new StringBuilder();
 
		int n=in.nextInt(),
			m=in.nextInt();

		ArrayList<Pair> fact =new ArrayList<>();
		ArrayList<Pair> fact0=new ArrayList<>();
		ArrayList<Pair> fact1=new ArrayList<>();

		for(int i=0; i<m; i++){
			int t=in.nextInt(),
				l=in.nextInt(),
				r=in.nextInt();
			Pair pair=new Pair(t, l, r);
			if(t==0)
				fact0.add(pair);
			else
				fact1.add(pair);
		}
		fact1=compress(fact1);

		for(Pair item: fact0){
			if(check(item, fact1))
				continue;
			else{
				System.out.println("NO");
				return;
			}
		}

		fact.addAll(fact1);
		fact.addAll(fact0);
		Collections.sort(fact);

		int a[]=new int[n+1];
		for(Pair item: fact){

			if(item.t==0){
				if(a[item.l]==0)
					a[item.l]=n;
				a[item.r]=a[item.l]-1;
			}
			else{
				int min=n;
				for(int i=item.l; i<=item.r; i++)
					if(a[i]!=0)
						min=Math.min(min, a[i]);
				for(int i=item.l; i<=item.r; i++)
					a[i]=min;
			}
		}

		System.out.println("YES");
		for(int i=1; i<=n; i++)
			System.out.print(Math.max(1, a[i])+" ");
	}
}