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;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int t, i, j, k, l;
  long long mod = 1e9 + 7;
  int n, m;
  cin >> n >> m;
  vector<int> a(n);
  vector<pair<int, int> > d, nd;
  while (m--) {
    int r;
    cin >> t >> l >> r;
    l--, r--;
    if (t)
      nd.push_back({l, r});
    else
      d.push_back({l, r + 1});
  }
  sort(nd.begin(), nd.end());
  for (i = 0; i + 1 < nd.size(); i++)
    if (nd[i].second >= nd[i + 1].first)
      nd[i].second = max(nd[i].second, nd[i + 1].second),
      nd.erase(nd.begin() + i + 1), i--;
  for (i = 0; i < nd.size(); i++) {
    for (j = nd[i].first; j < nd[i].second; j++) a[j] = 1;
    a[j] = 1e3 + 1;
  }
  for (i = 0, j = 1e3; i < n; i++) {
    if (a[i] > 0)
      j = 1e3;
    else
      a[i] = j--;
  }
  k = 0;
  for (auto x : d) k += is_sorted(a.begin() + x.first, a.begin() + x.second);
  if (k)
    cout << "NO\n";
  else {
    cout << "YES\n";
    for (auto x : a) cout << x << " ";
    cout << endl;
  }
  return 0;
}