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;
int n, m;
int b[1005];
int c[1005];
int d[1005];
int l[1005];
int r[1005];
pair<pair<int, int>, int> a[1005];
int main() {
  cin.tie(0);
  cout.tie(0);
  ios_base::sync_with_stdio(0);
  cin >> n >> m;
  int i, j;
  for (i = 1; i <= m; i++) {
    int t, l, r;
    cin >> a[i].second >> a[i].first.first >> a[i].first.second;
  }
  sort(a + 1, a + m + 1);
  for (i = 1; i <= m; i++) {
    int t = a[i].second;
    l[i] = a[i].first.first;
    r[i] = a[i].first.second;
    if (t == 1) {
      for (j = l[i] - 1; j >= 1; j--) {
        if (b[j] >= l[i]) b[j] = max(b[j], r[i]);
      }
      for (j = r[i]; j >= l[i]; j--) {
        b[j] = max(b[j], r[i]);
      }
    }
    if (t == 0) {
      for (j = l[i]; j >= 1; j--) {
        if (c[j] == 0)
          c[j] = r[i];
        else
          c[j] = min(c[j], r[i]);
      }
    }
  }
  for (i = 1; i <= n; i++) {
    if (c[i] != 0 && c[i] <= b[i]) {
      cout << "NO";
      return 0;
    }
  }
  for (i = 1; i <= n; i++) {
    for (j = i + 1; j <= b[i]; j++) {
      d[j] = 1;
    }
  }
  int cnt = 0;
  for (i = 1; i <= n; i++) {
    for (j = i + 1; j <= c[i]; j++) {
      if (d[j] == 0) {
        d[j] = --cnt;
      }
    }
  }
  for (i = 1; i <= n; i++) d[i] -= cnt - 1;
  cout << "YES\n";
  for (i = 1; i <= n; i++) cout << d[i] << " ";
  return 0;
}