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 = 1e3 + 15;
int dsu[N];
int f(int x) {
  if (dsu[x] == x) {
    return x;
  }
  return dsu[x] = f(dsu[x]);
}
void merge(int x, int y) {
  x = f(x), y = f(y);
  if (x != y) {
    dsu[x] = y;
  }
}
int val[N];
void solve() {
  int n, m;
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= n; ++i) {
    dsu[i] = i;
  }
  vector<pair<int, int> > v;
  for (int i = 0; i < m; ++i) {
    int l, r, t;
    scanf("%d %d %d", &t, &l, &r);
    if (t == 1) {
      for (int j = l + 1; j <= r; ++j) {
        merge(l, j);
      }
    } else {
      v.push_back(make_pair(l, r));
    }
  }
  vector<pair<int, int> > seg;
  for (int i = 1; i <= n;) {
    int j;
    for (j = i; j <= n; ++j) {
      if (f(j) != f(i)) {
        break;
      }
    }
    seg.push_back(make_pair(i, j - 1));
    i = j;
  }
  int maxi = 0;
  for (int i = (int)seg.size() - 1; i >= 0; --i) {
    int l = seg[i].first, r = seg[i].second;
    for (int j = l; j <= r; ++j) {
      val[j] = ++maxi;
    }
  }
  for (int i = 0; i < (int)v.size(); ++i) {
    int l = v[i].first, r = v[i].second;
    bool isSorted = true;
    for (int j = l + 1; j <= r; ++j) {
      if (val[j - 1] > val[j]) {
        isSorted = false;
      }
    }
    if (isSorted) {
      printf("NO\n");
      return;
    }
  }
  printf("YES\n");
  for (int i = 1; i <= n; ++i) {
    printf("%d ", val[i]);
  }
  cout << endl;
}
int main() { solve(); }