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.ArrayList;
import java.util.Collections;
import java.util.StringTokenizer;

public class Main1187C
{
    static ArrayList<Se>[] se;
    static int[] parent;
    static int N,M;
    public static void main(String[] args) throws IOException
    {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
        StringTokenizer st = new StringTokenizer(br.readLine());
        N = Integer.parseInt(st.nextToken());
        M = Integer.parseInt(st.nextToken());
        parent = new int[N+1];
        se = new ArrayList[2];
        for(int i=0;i<2;++i) se[i] = new ArrayList<>();
        for(int i=0;i<=N;++i) parent[i] = i;
        for(int i=0;i<M;++i)
        {
            st = new StringTokenizer(br.readLine());
            se[Integer.parseInt(st.nextToken())].add(new Se(Integer.parseInt(st.nextToken()),Integer.parseInt(st.nextToken())));
        }
        Collections.sort(se[1]);
        if(se[1].size()>0)
        {
            for (int i = se[1].get(0).s; i < se[1].get(0).e; ++i)
            {
                merge(i, i + 1);
            }
            for (int i = 1; i < se[1].size(); ++i)
            {
                int l, r = se[1].get(i).e;
                if (se[1].get(i).s < se[1].get(i - 1).e)
                {
                    l = se[1].get(i - 1).e;
                }
                else
                {
                    l = se[1].get(i).s;
                }
                for (; l < r; ++l)
                    merge(l, l + 1);
            }
        }
        for(Se s : se[0])
        {
            if(find(s.s)==find(s.e))
            {
                System.out.print("NO");
                return;
            }
        }
        int start = 10000;
        bw.write("YES\n"+start+" ");
        for(int i=2;i<=N;++i)
        {
            if(find(i-1)==find(i))
                bw.write(++start+" ");
            else
                bw.write(--start+" ");
        }
        bw.close();
    }

    static int find(int u)
    {
        if(parent[u]==u) return u;
        return parent[u]=find(parent[u]);
    }

    static void merge(int u, int v)
    {
        u=find(u); v=find(v);
        if(u==v) return;
        parent[u]=v;
    }
}
class Se implements Comparable<Se>
{
    int s,e;
    Se(int s, int e)
    {
        this.s = s; this.e = e;
    }
    public int compareTo(Se se)
    {
        return s==se.s?e-se.e:s-se.s;
    }
}