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 par[1005], size[1005];
int find_set(int v) {
  if (v == par[v]) return v;
  return par[v] = find_set(par[v]);
}
void make_set(int v) {
  par[v] = v;
  size[v] = 1;
}
void union_sets(int a, int b) {
  a = find_set(a);
  b = find_set(b);
  if (a != b) {
    if (size[a] < size[b]) swap(a, b);
    par[b] = a;
    size[a] += size[b];
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n, m, i, j, t, l, r;
  cin >> n >> m;
  for (i = 1; i <= n; i++) make_set(i);
  vector<pair<long long int, long long int> > v;
  vector<long long int> ans;
  for (i = 0; i < m; i++) {
    cin >> t >> l >> r;
    if (t == 0)
      v.push_back(make_pair(l, r));
    else {
      for (j = l; j <= r; j++) union_sets(l, j);
    }
  }
  for (auto j : v) {
    if (find_set(j.first) == find_set(j.second)) {
      cout << "NO";
      return 0;
    }
  }
  make_set(0);
  union_sets(0, 1);
  i = 1;
  long long nxt = 1000;
  long long st = 1000;
  cout << "YES\n";
  while (i <= n) {
    if (find_set(i) == find_set(i - 1)) {
      ans.push_back(st++);
      i++;
    } else {
      nxt--;
      st = nxt;
      ans.push_back(st++);
      i++;
    }
  }
  for (auto i : ans) cout << i << " ";
  return 0;
}