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;
long long mod = 1000000007LL;
long long large = 2000000000000000000LL;
vector<int> mini, addv;
int ul, ur, uv;
void update(int o, int l, int r) {
  if (ul <= l && ur >= r) {
    addv[o] += uv;
    mini[o] += uv;
  } else {
    int m = (l + r) / 2;
    if (ul <= m) update(o * 2, l, m);
    if (ur > m) update(o * 2 + 1, m + 1, r);
    mini[o] = min(mini[o * 2], mini[o * 2 + 1]) + addv[o];
  }
}
int ql, qr, qv;
void query(int o, int l, int r, int add) {
  if (ql <= l && qr >= r) {
    qv = min(qv, mini[o] + add);
  } else {
    int m = (l + r) / 2;
    if (ql <= m) query(o * 2, l, m, add + addv[o]);
    if (qr > m) query(o * 2 + 1, m + 1, r, add + addv[o]);
  }
}
int main() {
  int n, m;
  cin >> n >> m;
  mini.assign(4 * n + 10, 0);
  addv = mini;
  vector<pair<pair<int, int>, int> > e;
  for (int i = 0; i < m; i++) {
    int l, r, v;
    cin >> v >> l >> r;
    l--;
    r--;
    r--;
    e.push_back(pair<pair<int, int>, int>(pair<int, int>(l, r), v));
  }
  for (int i = 0; i < m; i++) {
    if (e[i].second == 1) {
      ul = e[i].first.first;
      ur = e[i].first.second;
      uv = 1;
      update(1, 0, n - 2);
    }
  }
  for (int i = 0; i < m; i++) {
    if (e[i].second == 0) {
      ql = e[i].first.first;
      qr = e[i].first.second;
      qv = 1;
      query(1, 0, n - 2, 0);
      if (qv == 1) {
        cout << "NO" << endl;
        return 0;
      }
    }
  }
  cout << "YES" << endl;
  vector<int> a(n, 5000);
  for (int i = 1; i < n; i++) {
    ql = qr = i - 1;
    qv = 1;
    query(1, 0, n - 2, 0);
    if (qv == 1)
      a[i] = a[i - 1];
    else
      a[i] = a[i - 1] - 1;
  }
  for (int i = 0; i < n; i++) cout << a[i] << " ";
  cout << endl;
  return 0;
}