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 = 500010;
int l[N];
int r[N];
int p[N];
int q[N];
vector<pair<int, int> > g[N];
int t[N];
void upd(int n, int pos, int x) {
  for (; pos <= n; pos |= pos + 1) t[pos] += x;
}
int get(int pos) {
  int sum = 0;
  for (; pos > 0; pos &= pos + 1, pos--) sum += t[pos];
  return sum;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  vector<pair<int, int> > v;
  for (int i = 1; i <= n; i++)
    cin >> l[i] >> r[i], g[l[i]].push_back({r[i], i});
  set<pair<int, int> > st;
  for (int i = 1; i <= n; i++) {
    for (auto cur : g[i]) {
      st.insert(cur);
    }
    int ind = (*st.begin()).second;
    p[ind] = i;
    st.erase(st.begin());
    if (!st.empty()) {
      v.push_back({ind, (*st.begin()).second});
    }
  }
  for (auto cur : v) {
    int i = cur.first;
    int j = cur.second;
    if (p[j] >= l[i] && p[j] <= r[i] && p[i] >= l[j] && p[i] <= r[j]) {
      cout << "NO\n";
      for (int t = 1; t <= n; t++) cout << p[t] << " ";
      cout << '\n';
      swap(p[i], p[j]);
      for (int t = 1; t <= n; t++) cout << p[t] << " ";
      cout << '\n';
      return 0;
    }
  }
  cout << "YES\n";
  for (int t = 1; t <= n; t++) cout << p[t] << " ";
  cout << '\n';
  return 0;
}
