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 N = 1005;
int n, m;
vector<pair<int, int> > sorted, unsorted;
inline bool cmp(pair<int, int> x, pair<int, int> y) {
  if (x.first != y.first) return x.first < y.first;
  return x.second < y.second;
}
vector<pair<int, int> > v;
bool vis[N];
int idx[N];
int a[N];
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) {
    int t, l, r;
    scanf("%d%d%d", &t, &l, &r);
    if (t == 1)
      sorted.push_back(pair<int, int>(l, r));
    else
      unsorted.push_back(pair<int, int>(l, r));
  }
  sort(sorted.begin(), sorted.end(), cmp);
  for (int i = 0; i < sorted.size(); i++) {
    pair<int, int> cur = sorted[i];
    if (v.empty() || v.back().second < cur.first)
      v.push_back(cur);
    else {
      if (v.back().second < cur.second) v[v.size() - 1].second = cur.second;
    }
  }
  for (int i = 0; i < unsorted.size(); i++)
    for (int j = 0; j < v.size(); j++) {
      if (v[j].first <= unsorted[i].first &&
          v[j].second >= unsorted[i].second) {
        printf("NO\n");
        return 0;
      }
    }
  printf("YES\n");
  for (int i = 0; i < v.size(); i++)
    for (int j = v[i].first; j <= v[i].second; j++) vis[j] = true;
  for (int i = 1; i <= n; i++)
    if (!vis[i]) v.push_back(pair<int, int>(i, i));
  sort(v.begin(), v.end(), cmp);
  for (int i = 0; i < v.size(); i++) idx[i] = i;
  int maxv = idx[v.size() - 1];
  sort(unsorted.begin(), unsorted.end(), cmp);
  for (int i = unsorted.size() - 1; i >= 0; i--) {
    bool flag = false;
    for (int j = 0; j < v.size() - 1 && !flag; j++)
      if (unsorted[i].first < v[j + 1].first &&
          unsorted[i].second >= v[j + 1].first)
        if (idx[j] < idx[j + 1]) {
          idx[j] = ++maxv;
          flag = true;
        }
  }
  for (int i = 0; i < v.size(); i++)
    for (int j = v[i].first; j <= v[i].second; j++) a[j] = idx[i] + 1;
  for (int i = 1; i <= n; i++) printf("%d ", a[i]);
  return 0;
}