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;
const int oo = 1e9 + 9;
const long long inf = 1e18 + 18;
const int max6 = 1e6 + 6;
const int modx = 1e9 + 7;
const int mody = 997;
const int base = 137;
int a[max6], dd[max6];
int main() {
  ios_base::sync_with_stdio(false);
  int n, m;
  cin >> n >> m;
  for (int i = 1; i <= n; ++i) a[i] = 2 * n - i;
  vector<pair<int, pair<int, int> > > waiting;
  set<pair<int, int> > one;
  for (int i = 1; i <= m; ++i) {
    int type, l, r;
    cin >> type >> l >> r;
    waiting.push_back({type, {l, r}});
    if (type) one.insert({l, r});
  }
  for (auto& v : one) {
    int l = v.first;
    int r = v.second;
    for (int i = l + 1; i <= r; ++i) a[i] = max(a[i], a[i - 1]);
  }
  for (auto& v : waiting) {
    int type = v.first;
    int l = v.second.first;
    int r = v.second.second;
    bool ok = true;
    for (int i = l + 1; i <= r; ++i) {
      if (type && a[i] < a[i - 1]) ok = false;
      if (type == 0 && a[i] < a[i - 1]) ok = false;
    }
    if (type == 1 && ok) continue;
    if (type == 0 && !ok) continue;
    cout << "NO";
    return 0;
  }
  cout << "YES\n";
  for (int i = 1; i <= n; ++i) cout << a[i] << " ";
}