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() {
  int n;
  scanf("%d", &n);
  vector<int> l(n), r(n);
  vector<pair<int, int>> st[n];
  for (int i = 0; i < n; i++) {
    scanf("%d%d", &l[i], &r[i]), l[i]--, r[i]--;
    st[l[i]].push_back({r[i], i});
  }
  set<pair<int, int>> sii;
  pair<int, int> prs = {-1, -1};
  vector<int> ans(n), dq;
  for (int i = 0; i < n; i++) {
    for (auto& p : st[i]) sii.insert(p);
    int x = sii.begin()->second;
    sii.erase(sii.begin());
    ans[x] = i;
    while (!dq.empty() && r[dq.back()] < i) dq.pop_back();
    if (!dq.empty() && l[x] <= ans[dq.back()]) prs = {dq.back(), x};
    dq.push_back(x);
  }
  printf(~prs.first ? "NO\n" : "YES\n");
  for (int& i : ans) printf("%d ", i + 1);
  printf("\n");
  if (prs.first == -1) return 0;
  swap(ans[prs.first], ans[prs.second]);
  for (int& i : ans) printf("%d ", i + 1);
  printf("\n");
}