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 mxn = 200000;
int n;
int a[mxn], b[mxn], first[mxn], ff[mxn];
vector<int> v[mxn];
auto cmp = [](int x, int y) { return b[x] > b[y]; };
priority_queue<int, vector<int>, decltype(cmp)> pq(cmp);
map<int, int> second;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i] >> b[i];
    a[i]--, b[i]--;
    v[a[i]].push_back(i);
  }
  int p = -1, q = -1;
  for (int i = 0; i < n; i++) {
    for (int j : v[i]) pq.push(j);
    int x = pq.top();
    pq.pop();
    first[i] = x, ff[x] = i + 1;
    if (!~p) {
      auto it = second.lower_bound(i);
      if (it != second.end() && it->second >= a[x]) {
        p = first[it->second], q = x;
      } else {
        it = second.insert({b[x], i}).first;
        if (it->second <= i) {
          it->second = i;
          if (next(it) != second.end() && i <= next(it)->second)
            it = second.erase(it);
          while (it != second.begin() && (--it)->second <= i)
            it = second.erase(it);
        }
      }
    }
  }
  cout << (!~p ? "YES" : "NO") << '\n';
  for (int i = 0; i < 1 + !!~p; i++) {
    cout << ff[0];
    for (int i = 1; i < n; i++) cout << " " << ff[i];
    cout << '\n';
    if (~p) swap(ff[p], ff[q]);
  }
  return 0;
}
