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 = 200100;
int n, a[N], b[N], p[N], pa[N];
vector<pair<int, int>> g[N];
void output() {
  static int out[N];
  for (int i = 1; i <= n; ++i) {
    out[p[i]] = i;
  }
  for (int i = 1; i <= n; ++i) {
    cout << out[i] << " \n"[i == n];
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i] >> b[i];
    g[a[i]].emplace_back(b[i], i);
  }
  set<pair<int, int>> st;
  for (int i = 1; i <= n; ++i) {
    st.insert(g[i].begin(), g[i].end());
    p[i] = st.begin()->second;
    st.erase(st.begin());
  }
  set<int> ps;
  for (int i = 1; i <= n; ++i) ps.insert(i);
  for (int i = 1; i <= n; ++i) {
    if (!ps.count(i)) continue;
    function<bool(int)> dfs;
    dfs = [&](int u) -> bool {
      int ut = p[u], v;
      for (auto it = ps.lower_bound(a[ut]);
           it != ps.end() && (v = *it) <= b[ut]; it = ps.upper_bound(v)) {
        if (v == u) continue;
        if (pa[v]) {
          cout << "NO\n";
          output();
          for (; u != v; u = pa[u]) {
            p[u] = p[pa[u]];
          }
          p[v] = ut;
          output();
          return true;
        }
        pa[v] = u;
        if (dfs(v)) return true;
        pa[v] = 0;
        ps.erase(v);
      }
      return false;
    };
    pa[i] = i;
    if (dfs(i)) return 0;
    pa[i] = 0;
    ps.erase(i);
  }
  cout << "YES\n";
  output();
  return 0;
}