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 CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long a, b, c, n, m, l, r, open1, open2, pos = 500000000, otv[2000],
                                             ans[2000];
vector<long long> let;
multiset<pair<pair<int, int>, int>> lol;
multiset<pair<long long, long long>> lol1, lol2, lol3;
string s;
int main() {
  ans[0] = 500000000;
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    cin >> a >> l >> r;
    if (a == 1) {
      for (auto it = lol1.begin(); it != lol1.end(); it++) {
        if (max(l, (*it).first) <= min(r, (*it).second)) {
          l = min(l, (*it).first);
          r = max(r, (*it).second);
        } else
          lol3.insert(*it);
      }
      lol3.insert(make_pair(l, r));
      lol1 = lol3;
      lol3.clear();
    } else
      lol2.insert(make_pair(l, r));
  }
  for (auto it = lol1.begin(); it != lol1.end(); it++) {
    for (int i = (*it).first; i <= (*it).second; i++) {
      otv[i] = pos;
    }
    pos -= 10000;
  }
  pos = 999999999;
  for (int i = 1; i <= n; i++) {
    if (otv[i] == 0) {
      otv[i] = pos;
      pos--;
    } else
      pos = otv[i] - 1;
  }
  for (auto it = lol2.begin(); it != lol2.end(); it++) {
    bool aa = false;
    for (int i = (*it).first + 1; i <= (*it).second; i++) {
      if (otv[i] < otv[i - 1]) aa = true;
    }
    if (aa == false) {
      cout << "NO";
      return 0;
    }
  }
  cout << "YES\n";
  for (int i = 1; i <= n; i++) {
    cout << otv[i] + 1 << " ";
  }
}