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){
        Scanner sc=new Scanner(System.in);
        int n=sc.nextInt();
        int m=sc.nextInt();
        int[] arr=new int[n];//对应间隔
        ArrayList<int[]> listA=new ArrayList<>();
        ArrayList<int[]> listB=new ArrayList<>();
        
        for(int i=0;i<m;i++){
            int f=sc.nextInt();
            int s=sc.nextInt();
            int e=sc.nextInt();
            if(f==1) listA.add(new int[]{s-1,e-1});
            else listB.add(new int[]{s-1,e-1});
        }
       
        for(int[] A:listA){
            int s=A[0];
            int e=A[1];
            arr[s]++;
            arr[e]--;
        }
        int cnt=0;
        int c=1;
        for(int i=0;i<n;i++){
            cnt+=arr[i];
            if(cnt>0){
                arr[i]=c;
            }else{
                c++;
                arr[i]=0;
            }
        }
        
        for(int[] B:listB){
            int s=B[0];
            int e=B[1]-1;
            // System.out.println(arr[s]+",,,"+arr[e-1]);
            if(s<=e&&arr[s]==arr[e]&&arr[s]!=0){
                System.out.println("NO");
                return;
            }
        }
        int cur=10000;
        System.out.println("YES");
        System.out.println(cur);
        for(int i=0;i<n-1;i++){
            if(arr[i]>0){
                cur++;
            }else{
                cur--;
            }
            System.out.println(cur);
        }
    }
}