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;
const int maxn = 2e5 + 10;
int n, m;
vector<pair<int, int> > g[2];
int a[maxn];
int ans[maxn];
int vis[maxn];
int main() {
  cin >> n >> m;
  int t, l, r;
  for (int i = 1; i <= m; i++) {
    cin >> t >> l >> r;
    g[t].push_back({l, r});
  }
  for (int i = 0; i < g[1].size(); i++) {
    l = g[1][i].first, r = g[1][i].second;
    a[l]++;
    a[r]--;
  }
  vector<int> pre0(n + 5, 0);
  for (int i = 1; i <= n; i++) {
    a[i] += a[i - 1];
    pre0[i] = pre0[i - 1] + (a[i] == 0);
  }
  for (int i = 0; i < g[0].size(); i++) {
    int l = g[0][i].first, r = g[0][i].second;
    if (pre0[r - 1] - pre0[l - 1] == 0) {
      printf("NO\n");
      return 0;
    }
  }
  printf("YES\n");
  ans[1] = 2000;
  for (int i = 2; i <= n; i++) {
    if (a[i - 1]) {
      ans[i] = ans[i - 1];
    } else {
      ans[i] = ans[i - 1] - 1;
    }
  }
  for (int i = 1; i <= n; i++) {
    printf("%d ", ans[i]);
  }
  return 0;
}