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 PYTHON3):

Solution:
"""
array of length n,
you have m facts,
truple ti,li ri
t=1 means sub array from l-r is sorted in non-dec
t=0 means sub array is not sorted


"""

n,m= map(int,input().split())
A=[0]*n
R=[]
for t in range(m):
    t,L,r= map(int,input().split())
    if(t==1):
        for i in range(L,r):
            A[i]=1
    else:
        R.append([L,r])
valid=True
for exp in R:
    if not(0 in A[exp[0]:exp[1]]):
        if not(-1 in A[exp[0]:exp[1]]):
            valid=False
            break
    else:
        A[A[exp[0]:exp[1]].index(0)+exp[0]]=-1

else:
    print("YES")
    ans=[]
    Ld=3000
    for i in A:
        if(i==0):
            ans.append(str(Ld))
        elif(i==1):
            ans.append(str(Ld+1))
            Ld+=1
        else:
            ans.append(str(Ld-1))
            Ld-=1
    print(" ".join(ans))
if not (valid):
    print("NO")