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 N = 2e5 + 5;
vector<int> l[N], r[N];
int L[N], R[N], ans[N], pos[N];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> L[i] >> R[i];
    l[L[i]].push_back(i);
    r[R[i] + 1].push_back(i);
  }
  pair<int, int> w = {-1, -1};
  set<pair<int, int> > st;
  set<int> z;
  for (int i = 1; i <= n; i++) {
    for (auto it : r[i]) {
      z.erase(pos[it]);
    }
    for (auto it : l[i]) {
      st.insert({R[it], it});
    }
    auto it = *st.begin();
    ans[i] = it.second;
    pos[it.second] = i;
    st.erase(st.begin());
    if (z.size() && *z.rbegin() >= L[it.second]) {
      w = {ans[*z.rbegin()], it.second};
    }
    z.insert(i);
  }
  cout << (w.first == -1 ? "YES\n" : "NO\n");
  for (int i = 1; i <= n; i++) cout << pos[i] << ' ';
  if (w.first != -1) {
    cout << "\n";
    swap(pos[w.first], pos[w.second]);
    for (int i = 1; i <= n; i++) cout << pos[i] << ' ';
  }
  return 0;
}
