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;
const int N = 200005;
int n, af[N], l[N], r[N], a[N], pos[N];
vector<pair<int, int>> eve[N];
priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>>
    pq;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> l[i] >> r[i];
    eve[l[i]].push_back({r[i], i});
  }
  for (int i = 1; i <= n; i++) {
    for (pair<int, int>& cur : eve[i]) {
      pq.push(cur);
    }
    auto [r, ind] = pq.top();
    pq.pop();
    a[ind] = i;
    pos[i] = ind;
    if (!pq.empty()) {
      af[ind] = pq.top().second;
    }
  }
  for (int i = 1; i <= n; i++) {
    int cur = i, nxt = af[i];
    if (nxt != 0 && l[cur] <= a[nxt] && a[nxt] <= r[cur] && l[nxt] <= a[cur] &&
        a[cur] <= r[nxt]) {
      cout << "NO\n";
      for (int i = 1; i <= n; i++) {
        cout << a[i] << " \n"[i == n];
      }
      swap(a[cur], a[nxt]);
      for (int i = 1; i <= n; i++) {
        cout << a[i] << " \n"[i == n];
      }
      return 0;
    }
  }
  cout << "YES\n";
  for (int i = 1; i <= n; i++) {
    cout << a[i] << " \n"[i == n];
  }
}