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;
class DSU {
 public:
  vector<int> Parent, Rank, Size;
  void UnionFindInit(int N) {
    Parent.resize(N);
    for (int x = 0; x < N; x++) Parent[x] = x;
    Rank.resize(N);
    Size.resize(N, 1);
  }
  int FindParent(int x) {
    return (Parent[x] == x) ? x : Parent[x] = FindParent(Parent[x]);
  }
  void MergeUnion(int x, int y) {
    x = FindParent(x);
    y = FindParent(y);
    if (x == y) return;
    if (Rank[x] >= Rank[y]) {
      if (Rank[x] == Rank[y]) Rank[x]++;
      Parent[y] = Parent[x];
      Size[x] += Size[y];
    } else {
      Parent[x] = Parent[y];
      Size[y] += Size[x];
    }
    return;
  }
};
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  int N, M;
  cin >> N >> M;
  DSU UF;
  UF.UnionFindInit(N);
  vector<pair<int, int> > S;
  while (M--) {
    int t, l, r;
    cin >> t >> l >> r;
    l--, r--;
    if (!t)
      S.push_back({l, r});
    else
      for (int x = l + 1; x <= r; x++) UF.MergeUnion(l, x);
  }
  bool flag = true;
  vector<int> A(N);
  vector<bool> Vis(N, 0);
  for (auto &i : S) {
    if (UF.FindParent(i.first) == UF.FindParent(i.second)) {
      flag = false;
      break;
    }
    while (UF.FindParent(i.first) != UF.FindParent(i.second)) i.second--;
    i.second++;
    Vis[i.second - 1] = 1;
  }
  int curr = 1e5;
  for (int x = 0; x < N; x++) {
    A[x] = curr;
    if (Vis[x]) curr--;
  }
  if (flag) {
    cout << "YES" << endl;
    for (auto i : A) cout << i << " ";
  } else
    cout << "NO";
  cout << endl;
  return 0;
}