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 int MX = 200005;
int n, a[MX], b[MX], res[MX], ot[MX], f[MX];
vector<long long> pos[MX], del[MX];
set<pair<int, int> > st;
set<int> in;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i] >> b[i];
    pos[a[i]].push_back(i);
  }
  for (int i = 1; i <= n; i++) {
    for (int j : pos[i]) st.emplace(b[j], j);
    f[i] = st.begin()->second;
    res[f[i]] = i;
    st.erase(st.begin());
  }
  for (int i = 1; i <= n; i++) {
    if (in.size() && *in.rbegin() >= a[f[i]]) {
      for (int i = 0; i < n; i++) ot[i] = res[i];
      swap(ot[f[i]], ot[f[*in.rbegin()]]);
      cout << "NO" << '\n';
      for (int i = 0; i < n; i++) cout << res[i] << " ";
      cout << '\n';
      for (int i = 0; i < n; i++) cout << ot[i] << " ";
      cout << '\n';
      return 0;
    }
    in.insert(i);
    del[b[f[i]]].push_back(i);
    for (int j : del[i]) in.erase(j);
  }
  cout << "YES" << '\n';
  for (int i = 0; i < n; i++) cout << res[i] << " ";
  cout << '\n';
  return 0;
}
