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;
inline void pisz(int n) { printf("%d\n", n); }
const int dr[4] = {-1, 0, 1, 0};
const int dc[4] = {0, -1, 0, 1};
const int drr[8] = {-1, -1, 0, 1, 1, 1, 0, -1};
const int dcc[8] = {0, -1, -1, -1, 0, 1, 1, 1};
template <typename T, typename TT>
ostream& operator<<(ostream& s, pair<T, TT> t) {
  return s << "(" << t.first << "," << t.second << ")";
}
template <typename T>
ostream& operator<<(ostream& s, vector<T> t) {
  for (int i = 0; i < ((int)((t).size())); i++) s << t[i] << " ";
  return s;
}
class UnionFind {
 private:
  vector<int> rank, p;

 public:
  UnionFind(int n) {
    for (int i = 0; i < n; i++) p.push_back(i);
    rank.assign(n, 0);
  }
  int findSet(int s) { return (p[s] == s) ? s : (p[s] = findSet(p[s])); }
  bool isSameSet(int i, int j) { return findSet(i) == findSet(j); }
  void unionSet(int i, int j) {
    if (isSameSet(i, j)) return;
    int x = findSet(i), y = findSet(j);
    if (rank[x] > rank[y])
      p[y] = x;
    else {
      p[x] = y;
      if (rank[x] == rank[y]) rank[y]++;
    }
  }
};
bool fcomp(double A, double B) {
  double EPSILON = numeric_limits<double>::epsilon();
  double diff = A - B;
  return fabs(diff) < EPSILON;
}
double x_y(int x1, int y1, int x2, int y2) {
  return sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2));
}
int main() {
  int(n), (m);
  scanf("%d %d", &(n), &(m));
  vector<pair<int, int> > q;
  int rng[n + 1];
  memset(rng, -1, sizeof(rng));
  for (int i = 0; i < m; i++) {
    int(t), (l), (r);
    scanf("%d %d %d", &(t), &(l), &(r));
    if (t) {
      rng[l] = (rng[l] < r ? r : rng[l]);
    } else
      q.push_back({l, r});
  }
  for (int i = 1; i < n + 1; i++)
    rng[i] = (rng[i] < rng[i - 1] ? rng[i - 1] : rng[i]);
  int cur = 1e7;
  vector<int> ans(n + 1, -1);
  vector<int> dn(n + 1, 0);
  vector<pair<int, int> > sorted;
  for (int i = 1; i < n + 1; i++) {
    if (dn[i]) continue;
    if (rng[i] < i) continue;
    int l = i, r = i, p = cur;
    while (rng[r] != r) r = rng[r];
    sorted.push_back({l, r});
    for (int j = l; j <= r; j++) {
      dn[j] = 1;
      ans[j] = p++;
    }
    cur -= 1010;
  }
  int nw = 1e8, id = 1;
  while (ans[id] == -1) {
    ans[id] = nw--;
    id++;
  }
  sorted.push_back({n + 1, n + 1});
  cur = 1e7 - 1;
  for (int i = 1; i < sorted.size(); i++) {
    int p = cur;
    for (int j = sorted[i - 1].second + 1; j < sorted[i].first; j++)
      ans[j] = p--;
    cur -= 1010;
  }
  int ys = 1;
  for (int i = 0; i < q.size(); i++) {
    for (int j = 0; j < sorted.size(); j++) {
      if (sorted[j].first <= q[i].first && sorted[j].second >= q[i].second)
        ys = 0;
    }
  }
  if (!ys) {
    cout << "NO" << endl;
  } else {
    cout << "YES" << endl;
    for (int i = 1; i < ans.size(); i++) cout << ans[i] << " ";
    cout << endl;
  }
  return 0;
}