Problem: Phoenix is trying to take a photo of his n friends with labels 1, 2, ..., n who are lined up in a row in a special order. But before he can take the photo, his friends get distracted by a duck and mess up their order.

Now, Phoenix must restore the order but he doesn't remember completely! He only remembers that the i-th friend from the left had a label between a_i and b_i inclusive. Does there exist a unique way to order his friends based of his memory? 

Input

The first line contains one integer n (1 ≤ n ≤ 2⋅10^5) — the number of friends.

The i-th of the next n lines contain two integers a_i and b_i (1 ≤ a_i ≤ b_i ≤ n) — Phoenix's memory of the i-th position from the left.

It is guaranteed that Phoenix's memory is valid so there is at least one valid ordering.

Output

If Phoenix can reorder his friends in a unique order, print YES followed by n integers — the i-th integer should be the label of the i-th friend from the left.

Otherwise, print NO. Then, print any two distinct valid orderings on the following two lines. If are multiple solutions, print any.

Examples

Input


4
4 4
1 3
2 4
3 4


Output


YES
4 1 2 3 


Input


4
1 3
2 4
3 4
2 3


Output


NO
1 3 4 2 
1 2 4 3 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 7;
int perm[N];
pair<int, int> v[N];
vector<pair<int, int>> s[N];
signed main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> v[i].first >> v[i].second;
    s[v[i].first].push_back({v[i].second, i});
  }
  set<pair<int, int>> best;
  vector<pair<int, int>> ch;
  for (int i = 1; i <= n; i++) {
    for (auto elem : s[i]) best.insert(elem);
    assert(!best.empty());
    int p = (*best.begin()).second;
    perm[p] = i;
    best.erase(best.begin());
    if (!best.empty()) {
      ch.push_back({p, (*best.begin()).second});
    }
  }
  pair<int, int> res = {-1, -1};
  for (auto [a, b] : ch) {
    if (v[b].first <= perm[a] && perm[a] <= v[b].second) {
      if (v[a].first <= perm[b] && perm[b] <= v[a].second) {
        res = {a, b};
        break;
      }
    }
  }
  if (res.first == -1) {
    cout << "YES" << endl;
    for (int i = 1; i <= n; i++) cout << perm[i] << ' ';
    cout << endl;
  } else {
    cout << "NO" << endl;
    for (int i = 1; i <= n; i++) cout << perm[i] << ' ';
    cout << endl;
    swap(perm[res.first], perm[res.second]);
    for (int i = 1; i <= n; i++) cout << perm[i] << ' ';
    cout << endl;
  }
  return 0;
}