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;
using pii = pair<int, int>;
int A[200009], B[200009], ans[200009], R[200009];
vector<int> Q[200009];
int main() {
  int N;
  scanf("%d", &N);
  for (int i = 1; i <= N; i++) {
    scanf("%d%d", &A[i], &B[i]);
    Q[A[i]].push_back(i);
  }
  priority_queue<pii> pq;
  for (int i = 1; i <= N; i++) {
    for (auto& it : Q[i]) pq.push({-B[it], it});
    int x = pq.top().second;
    pq.pop();
    ans[x] = i;
    R[i] = x;
  }
  int x = -1, y = -1;
  set<int> S = {-1, N + 1};
  for (int i = 1; i <= N; i++) {
    for (auto& it : Q[i]) {
      pq.push({-B[it], it});
      S.insert(ans[it]);
    }
    auto it = S.find(i);
    if (*next(it) <= B[R[i]]) x = R[i], y = R[*next(it)];
    if (A[R[i]] <= *prev(it)) x = R[i], y = R[*prev(it)];
    while (pq.size() && -pq.top().first == i) {
      S.erase(ans[pq.top().second]);
      pq.pop();
    }
  }
  if (x == -1) {
    puts("YES");
    for (int i = 1; i <= N; i++) printf("%d ", ans[i]);
  } else {
    puts("NO");
    for (int i = 1; i <= N; i++) printf("%d ", ans[i]);
    puts("");
    swap(ans[x], ans[y]);
    for (int i = 1; i <= N; i++) printf("%d ", ans[i]);
  }
  return 0;
}