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;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int N;
  cin >> N;
  vector<int> L(N), R(N);
  vector<vector<int>> adds(N);
  for (int i = 0; i < N; ++i) {
    cin >> L[i] >> R[i];
    --L[i], --R[i];
    adds[L[i]].emplace_back(i);
  }
  multiset<pair<int, int>> ms;
  auto print = [&](vector<int> ord) {
    vector<int> perm(N);
    for (int i = 0; i < N; ++i) perm[ord[i]] = i;
    for (int v : perm) cout << v + 1 << " ";
    cout << "\n";
  };
  vector<int> ord(N);
  for (int l = 0; l < N; ++l) {
    for (int i : adds[l]) ms.emplace(R[i], i);
    ord[l] = ms.begin()->second;
    ms.erase(ms.begin());
  }
  vector<int> perm(N);
  for (int i = 0; i < N; ++i) perm[ord[i]] = i;
  vector<int> last(N);
  for (int i = 0; i < N; ++i) {
    if (i == 0 || R[ord[i - 1]] == i - 1) {
      last[i] = i;
    } else {
      last[i] = last[i - 1];
    }
  }
  for (int l = 0; l < N; ++l) {
    for (int i : adds[l]) ms.emplace(R[i], i);
    if (ms.size() >= 2) {
      int x = ms.begin()->second;
      int y = next(ms.begin())->second;
      if (last[perm[y]] <= R[x]) {
        vector<int> neword = ord;
        neword[l] = y;
        ms.erase(next(ms.begin()));
        for (++l; l < N; ++l) {
          for (int i : adds[l]) ms.emplace(R[i], i);
          neword[l] = ms.begin()->second;
          ms.erase(ms.begin());
        }
        cout << "NO\n";
        print(ord);
        print(neword);
        return 0;
      }
    }
    ms.erase(ms.begin());
  }
  cout << "YES\n";
  print(ord);
  return 0;
}