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 
#include <bits/stdc++.h>
using namespace std;
const long long INF = 2e9;
const long long MB = 31;
const long long MOD = 1e9 + 7;
void solve() {
  long long n;
  cin >> n;
  vector<pair<long long, long long>> a(n);
  for (long long i = 0; i < n; i++) {
    cin >> a[i].first >> a[i].second;
    a[i].first--;
    a[i].second--;
  }
  vector<vector<pair<long long, long long>>> b(n);
  for (long long i = 0; i < n; i++) {
    b[a[i].first].push_back({a[i].second, i});
  }
  vector<long long> id(n);
  set<pair<long long, long long>> st;
  for (long long i = 0; i < n; i++) {
    for (auto& p : b[i]) {
      st.insert(p);
    }
    auto p = *st.begin();
    st.erase(p);
    id[p.second] = i;
  }
  vector<long long> add(n);
  for (long long i = 0; i < n; i++) add[id[i]] = i;
  vector<vector<long long>> del(n);
  for (long long i = 0; i < n; i++)
    if (a[i].second + 1 < n) del[a[i].second + 1].push_back(i);
  st.clear();
  for (long long i = 0; i < n; i++) {
    for (long long j : del[i]) {
      st.erase({id[j], j});
    }
    if (!st.empty() && st.rbegin()->first >= a[add[i]].first) {
      long long ind = st.rbegin()->second;
      cout << "NO\n";
      for (long long i : id) cout << i + 1 << " ";
      cout << '\n';
      swap(id[st.rbegin()->second], id[add[i]]);
      for (long long i : id) cout << i + 1 << " ";
      return;
    }
    st.insert({i, add[i]});
  }
  cout << "YES\n";
  for (long long i : id) cout << i + 1 << " ";
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cout.precision(12);
  cout << fixed;
  long long t = 1;
  while (t--) solve();
}
