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>
#pragma GCC optimize("Ofast")
using namespace std;
int n, x, y;
int ans[200005];
vector<pair<int, int>> l[200005];
set<pair<int, int>> s;
pair<int, int> v[200005];
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> x >> y;
    l[x].push_back({y, i});
    v[i] = {x, y};
  }
  for (int i = 1; i <= n; i++) {
    for (auto &j : l[i]) s.insert(j);
    ans[s.begin()->second] = i;
    s.erase(s.begin());
  }
  bool ok = 0;
  for (int i = 1; i <= n; i++) {
    for (auto &j : l[i]) s.insert(j);
    if (s.size() > 1) {
      auto it = s.begin();
      it++;
      if (ans[it->second] <= s.begin()->first) {
        cout << "NO\n";
        for (int j = 1; j <= n; j++) cout << ans[j] << " ";
        cout << "\n";
        swap(ans[s.begin()->second], ans[it->second]);
        for (int j = 1; j <= n; j++) cout << ans[j] << " ";
        ok = 1;
        break;
      }
    }
    s.erase(s.begin());
  }
  if (!ok) {
    cout << "YES\n";
    for (int i = 1; i <= n; i++) cout << ans[i] << " ";
  }
  return 0;
}
