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;
vector<pair<long long, long long> > g[200005];
int xy[200005][2];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, u, v;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> u >> v;
    xy[i][0] = u;
    xy[i][1] = v;
    g[u].push_back(pair<long long, long long>(v, i));
  }
  priority_queue<pair<long long, long long>,
                 vector<pair<long long, long long> >,
                 greater<pair<long long, long long> > >
      pq;
  vector<int> res;
  vector<pair<long long, long long> > sw;
  for (int i = 1; i <= n; i++) {
    for (auto y : g[i]) pq.push(y);
    if (pq.top().first > i && pq.size() > 1) {
      pair<long long, long long> x = pq.top();
      pq.pop();
      sw.push_back(pair<long long, long long>(x.second, pq.top().second));
      pq.push(x);
    }
    res.push_back(pq.top().second);
    pq.pop();
  }
  if (!sw.size()) {
    cout << ("YES") << '\n';
    vector<int> rr(n);
    for (int i = 0; i < n; i++) rr[res[i]] = i + 1;
    for (auto y : rr) cout << y << ' ';
    cout << '\n';
  } else {
    vector<pair<long long, long long> > sw_ok;
    vector<int> rr(n);
    for (int i = 0; i < n; i++) rr[res[i]] = i + 1;
    for (auto y : sw)
      if (rr[y.second] <= xy[y.first][1]) sw_ok.push_back(y);
    if (sw_ok.size()) {
      cout << ("NO") << '\n';
      for (auto y : rr) cout << y << ' ';
      cout << '\n';
      for (int i = 0; i < n; i++) {
        if (res[i] == sw_ok[0].first)
          res[i] = sw_ok[0].second;
        else if (res[i] == sw_ok[0].second)
          res[i] = sw_ok[0].first;
      }
      for (int i = 0; i < n; i++) rr[res[i]] = i + 1;
      for (auto y : rr) cout << y << ' ';
      cout << '\n';
    } else {
      cout << ("YES") << '\n';
      for (auto y : rr) cout << y << ' ';
      cout << '\n';
    }
  }
  return 0;
}