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;
pair<int, int> Type0[1005], Type1[1005];
vector<pair<int, int> > V;
int Array[1005];
void PreProcess(int N, int m) {
  int i, j, l, r, start, end;
  for (i = 1; i <= N; ++i) Array[i] = N - i + 1;
  if (m == 0) return;
  sort(Type1 + 1, Type1 + m + 1);
  i = 2;
  j = 1;
  l = Type1[1].first;
  r = Type1[1].second;
  while (j <= N && i <= m) {
    if (j == l) {
      start = l;
      end = r;
    }
    if (j == r) {
      while (i <= m && Type1[i].second <= r) {
        ++i;
      }
      if (i > m)
        break;
      else if (Type1[i].first <= r)
        end = r = Type1[i].second;
      else {
        V.emplace_back(make_pair(start, end));
        l = Type1[i].first;
        r = Type1[i].second;
      }
      ++i;
    }
    ++j;
  }
  V.emplace_back(make_pair(l, r));
}
bool IsPossible(int N, int C0) {
  int i, j, l, r;
  for (i = 1; i <= C0; ++i) {
    l = Type0[i].first;
    r = Type0[i].second;
    for (j = 0; j < V.size(); ++j) {
      if (l >= V[j].first && r <= V[j].second) return false;
    }
  }
  return true;
}
void BuildArray(int N) {
  int i, start, end;
  for (i = 0; i < V.size(); ++i) {
    start = V[i].first;
    end = V[i].second;
    while (start < end) {
      swap(Array[start], Array[end]);
      ++start;
      --end;
    }
  }
}
int main() {
  int N, m, i, t, l, r;
  int C0 = 0, C1 = 0;
  scanf("%d%d", &N, &m);
  for (i = 1; i <= m; ++i) {
    scanf("%d%d%d", &t, &l, &r);
    if (t == 0) {
      ++C0;
      Type0[C0].first = l;
      Type0[C0].second = r;
    } else {
      ++C1;
      Type1[C1].first = l;
      Type1[C1].second = r;
    }
  }
  PreProcess(N, C1);
  if (IsPossible(N, C0) == false)
    puts("NO");
  else {
    BuildArray(N);
    puts("YES");
    for (i = 1; i <= N; ++i) printf("%d ", Array[i]);
    puts("");
  }
  return 0;
}