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.*;
public class Main {
    public static void main(String args[]) {
        // System.out.println("Hello World!");
        Scanner scn = new Scanner(System.in);
        int n = scn.nextInt();
        int[] arr = new int[n];
        int m = scn.nextInt();
        HashMap<Integer,Integer> nondec = new HashMap<>();
        HashMap<Integer,Integer> dec = new HashMap<>();
        ArrayList<pair> decr = new ArrayList();
        for(int i=0;i<m;++i){
            int t = scn.nextInt();
            int st = scn.nextInt();
            int en = scn.nextInt();
            if(t==0){
                decr.add(new pair(st-1,en-1));
            }else{
                for(int j=st;j<en;++j){
                    nondec.put(j,0);
                }
            }
        }
        Collections.sort(decr);
        for(int i=0;i<decr.size();++i){
            int st = decr.get(i).a;
            int en = decr.get(i).b;
            int flag = 0;
            for(int j=st+1;j<=en;++j){
                if(dec.containsKey(j)){
                    flag = 1;
                    break;
                }
            }
            if(flag==1){
                continue;
            }
            flag = 0;
            for(int j=st+1;j<=en;++j){
                if(nondec.containsKey(j)){
                    continue;
                }else{
                    dec.put(j,0);
                    flag = 1;
                }
            }
            if(flag==0){
                System.out.println("NO");
                System.exit(0);
            }




        }
        for(int i=0;i<n;++i){
            if(nondec.containsKey(i)){
                arr[i] = arr[i-1]+1;
            }
            else if(dec.containsKey(i)){
                arr[i] = arr[i-1]-1;
            }else{
                arr[i] = 1001;
            }
        }
        System.out.println("YES");
        for(int i=0;i<n;++i){
            System.out.print(arr[i]+" ");
        }
    }
 
    public static class pair implements Comparable<pair>{
        int a ;
        int b;
        pair(int m, int n){
            a = m;
            b = n;
        }
        public int compareTo(pair o){
            if(this.a!=o.a){
                return this.a-o.a;
            }
            return this.b-o.b;
        }
    }
}