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 = 1006;
struct node {
  int xy, l, r;
  friend bool operator<(node a, node b) { return a.l < b.l; }
} op[maxn];
int ans[maxn];
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; ++i) scanf("%d%d%d", &op[i].xy, &op[i].l, &op[i].r);
  sort(op + 1, op + 1 + m);
  int now = 1e8;
  for (int i = 1; i <= m; ++i) {
    if (op[i].xy == 1) {
      if (ans[op[i].l] == 0) {
        now -= 1000;
        ans[op[i].l] = now;
      }
      for (int j = op[i].l + 1; j <= op[i].r; ++j) {
        ans[j] = ans[j - 1];
      }
    }
  }
  now = 1e9;
  for (int i = 1; i <= n; ++i) {
    if (ans[i] == 0) {
      ans[i] = now - 1;
      now--;
    } else
      now = ans[i];
  }
  bool ok = true;
  for (int i = 1; i <= m; ++i) {
    if (op[i].xy == 0) {
      bool tp = false;
      for (int j = op[i].l; j < op[i].r; ++j) {
        if (ans[j] > ans[j + 1]) {
          tp = true;
          break;
        }
      }
      if (!tp) {
        ok = false;
        break;
      }
    }
  }
  if (!ok)
    printf("NO\n");
  else {
    printf("YES\n");
    for (int i = 1; i <= n; ++i) printf("%d ", ans[i]);
  }
  return 0;
}