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>
#pragma GCC optimize(2)
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
const long long mod = 998244353;
const long long INF = 1e9 + 7;
const int base = 131;
const double r = 0.57721566490153286060651209;
const double eps = 0.0000001;
int n, m;
int L[1010], R[1010], fa[1010], vis[1010];
int ans[1010];
struct node {
  int l, r, t;
} e[1010];
void init() {
  for (int i = 0; i < m; i++) {
    fa[i] = i;
  }
}
int find_root(int x) {
  if (x != fa[x]) fa[x] = find_root(fa[x]);
  return fa[x];
}
void merge_root(int x, int y) {
  int a = find_root(x);
  int b = find_root(y);
  if (a == b) return;
  fa[a] = b;
  L[b] = min(L[a], L[b]);
  R[b] = max(R[a], R[b]);
}
int main() {
  while (~scanf("%d%d", &n, &m)) {
    init();
    memset(vis, 0, sizeof(vis));
    ;
    for (int i = 0; i < m; i++) {
      scanf("%d%d%d", &e[i].t, &e[i].l, &e[i].r);
      L[i] = e[i].l, R[i] = e[i].r;
      if (e[i].t)
        for (int j = 0; j < i; j++) {
          if (e[j].t && ((e[i].l >= e[j].l && e[i].l <= e[j].r) ||
                         (e[i].r >= e[j].l && e[i].r <= e[j].r))) {
            merge_root(i, j);
          }
        }
    }
    vector<pair<int, int> > v;
    for (int i = 0; i < m; i++) {
      int f = find_root(i);
      if (vis[f] || !e[i].t) continue;
      vis[f] = 1;
      v.push_back(make_pair(L[f], R[f]));
    }
    sort(v.begin(), v.end());
    int now = 1e9 - 1000, flag = 0;
    for (int i = 0; i < m; i++) {
      if (e[i].t) continue;
      for (int j = 0; j < v.size(); j++) {
        if (v[j].first > e[i].r) break;
        if (e[i].l >= v[j].first && e[i].r <= v[j].second) {
          flag = 1;
          break;
        }
      }
      if (flag) break;
    }
    if (flag)
      printf("NO\n");
    else {
      printf("YES\n");
      for (int i = 1; i <= n; i++) ans[i] = n - i + 1;
      for (int i = 0; i < v.size(); i++) {
        for (int j = v[i].first + 1; j <= v[i].second; j++) {
          ans[j] = ans[j - 1];
        }
      }
      for (int i = 1; i <= n; i++) {
        printf("%d ", ans[i]);
      }
      printf("\n");
    }
  }
  return 0;
}