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;
vector<long long> primes;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  vector<pair<int, int> > zeros;
  vector<pair<int, int> > result;
  vector<pair<int, int> > ones;
  while (m--) {
    bool x;
    int a, b;
    cin >> x >> a >> b;
    if (!x)
      zeros.push_back(make_pair(a, b));
    else
      ones.push_back(make_pair(a, b));
  }
  sort(ones.begin(), ones.end());
  bool r(true);
  if (!ones.empty()) result.push_back(ones[0]);
  for (auto i : ones) {
    int X = result.size() - 1;
    if (result[X].first <= i.first && i.first <= result[X].second)
      result[X].second = max(result[X].second, i.second);
    else
      result.push_back(i);
  }
  for (auto i : zeros) {
    for (auto j : result) {
      if (j.first <= i.first && i.second <= j.second) {
        cout << "NO" << endl;
        return 0;
      }
    }
  }
  int d = n;
  vector<int> ans(n);
  for (auto i : result) {
    for (int j = i.first - 1; j < i.second - 1; ++j) ans[j] = 1;
    ans[i.second - 1] = 2;
  }
  for (int &i : ans) {
    if (i == 0)
      i = d--;
    else if (i == 2)
      i = d--;
    else if (i != 0)
      i = d;
  }
  cout << "YES" << endl;
  for (int i : ans) cout << i << " ";
  cout << endl;
  return 0;
}