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 = 1e6 + 10;
const int inf = 0x3f3f3f3f;
const int iinf = 1 << 30;
const long long linf = 2e18;
const int mod = 998244353;
const double eps = 1e-7;
template <class T>
void read(T &a) {
  int f = 1;
  a = 0;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    a = (a << 3) + (a << 1) + ch - '0';
    ch = getchar();
  }
  a *= f;
}
int n, m, tot;
int s[maxn], p[maxn];
struct node {
  int t, l, r;
} a[maxn], b[maxn];
signed main() {
  scanf("%d %d", &n, &m);
  for (long long i = (1); i <= (m); ++i)
    scanf("%d %d %d", &a[i].t, &a[i].l, &a[i].r);
  sort(a + 1, a + m + 1,
       [](node a, node b) { return a.t == b.t ? a.l < b.l : a.t > b.t; });
  for (long long i = (1); i <= (m); ++i) {
    if (a[i].t == 0)
      b[++tot] = a[i];
    else {
      if (a[i].l <= b[tot].r)
        b[tot].r = max(b[tot].r, a[i].r);
      else
        b[++tot] = a[i];
    }
  }
  sort(b + 1, b + tot + 1,
       [](node a, node b) { return a.l == b.l ? a.t > b.t : a.l < b.l; });
  int last = 0;
  for (long long i = (1); i <= (tot); ++i) {
    if (b[i].t == 0) {
      if (b[i].l <= last && b[i].r <= last) return puts("NO"), 0;
    } else
      last = b[i].r;
  }
  puts("YES");
  int now = 0, sum = 1000000;
  for (long long i = (1); i <= (tot); ++i) {
    if (b[i].t == 0) continue;
    while (now < b[i].l - 1) now++, printf("%d ", sum--);
    for (long long j = (b[i].l); j <= (b[i].r); ++j) now++, printf("%d ", sum);
    sum--;
  }
  while (now < n) now++, printf("%d ", sum--);
  return 0;
}