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:
n, m = map(int, input().split())
query1 = [0]*(n+1)
query0 = [0]*(n+1)
querylist0 = []
querylist1 = []
for _ in range(m):
    t, l, r = map(int, input().split())
    if t == 1:
        for i in range(l, r):
            query1[i] = 1
    else:
        querylist0.append([l, r-1])
        for i in range(l, r):
            query0[i] = 1

query1[0] = 2
query0[0] = 2
import itertools
now = 1
for key, group in itertools.groupby(query1):
    number = len(list(group))
    if key == 1:
        querylist1.append([now, now+number-1])
        now += number
    elif key == 0:
        now += number




flag = 1
for i in range(len(querylist1)):
    for j in range(len(querylist0)):
        if querylist1[i][0] <= querylist0[j][0] and querylist1[i][1] >= querylist0[j][1]:
            flag = 0



if flag == 0:
    print("NO")

else:
    print("YES")
    ans = []
    now = 10000
    ans.append(now)
    for i in range(1, n):
        if query1[i] == 1:
            now += 1
            ans.append(now)
        elif query0[i] == 1:
            now -= 1
            ans.append(now)
        else:
            now -= 1
            ans.append(now)

    print(" ".join(map(str, ans)))