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;
void no() { printf("NO\n"); }
void yes() { printf("YES\n"); }
const long long N = 1e5 + 5;
const long long M = 1e9 + 7;
void sol() {
  int n;
  int m;
  cin >> n;
  int con[n];
  vector<pair<int, int>> p;
  for (int i = 0; i < n; ++i) {
    con[i] = 0;
  }
  cin >> m;
  while (m--) {
    int x, y;
    int t;
    cin >> t;
    cin >> x >> y;
    --x;
    --y;
    if (t)
      con[x]++, con[y]--;
    else
      p.push_back({x, y});
  }
  int sum = 0;
  for (int i = 0; i + 1 < n; ++i) {
    sum += con[i];
    if (sum > 0)
      con[i] = 0;
    else
      con[i] = -1;
  }
  for (int i = 0; i < (int)p.size(); ++i) {
    bool ok = true;
    for (int j = p[i].first; j < p[i].second; ++j) {
      if (con[j] == -1) {
        ok = false;
        break;
      }
    }
    if (ok) {
      no();
      return;
    }
  }
  yes();
  sum = 1000000;
  cout << sum << " ";
  for (int i = 1; i < n; ++i) {
    if (con[i - 1] == 0)
      sum++;
    else
      sum--;
    cout << sum << " \n"[i + 1 == n];
  }
}
int main() {
  int tc = 1;
  while (tc--) {
    sol();
  }
}