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;
bool inotr(long long l, long long x, long long r) {
  return (x >= l) && (x < r);
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  pair<long long, long long> a[n];
  vector<pair<long long, long long> > b[n];
  for (long long i = 0; i < n; ++i) {
    cin >> a[i].first >> a[i].second;
    a[i].first--;
    b[a[i].first].push_back({a[i].second, i});
  }
  long long ans[n];
  long long u[n];
  set<pair<long long, pair<long long, long long> > > v;
  for (long long i = 0; i < n; ++i) {
    for (long long j = 0; j < b[i].size(); ++j) {
      pair<long long, long long> curr = b[i][j];
      v.insert({curr.first, {i, curr.second}});
    }
    pair<long long, pair<long long, long long> > h = (*v.begin());
    v.erase(h);
    ans[i] = h.second.second;
  }
  for (long long i = 0; i < n; ++i) {
    u[ans[i]] = i;
  }
  for (long long i = 0; i < (n - 1); ++i) {
    if (inotr(a[ans[i]].first, i + 1, a[ans[i]].second) &&
        inotr(a[ans[i + 1]].first, i, a[ans[i + 1]].second)) {
      cout << "NO" << endl;
      for (long long i = 0; i < n; ++i) {
        cout << u[i] + 1 << " ";
      }
      cout << endl;
      swap(ans[i], ans[i + 1]);
      for (long long i = 0; i < n; ++i) {
        u[ans[i]] = i;
      }
      for (long long i = 0; i < n; ++i) {
        cout << u[i] + 1 << " ";
      }
      return 0;
    }
  }
  for (long long i = 0; i < n; ++i) {
    sort(b[i].begin(), b[i].end());
    if (b[i].size() >= 2) {
      long long x = u[b[i][1].second];
      if (x < b[i][0].first) {
        cout << "NO" << endl;
        for (long long i = 0; i < n; ++i) {
          cout << u[i] + 1 << " ";
        }
        cout << endl;
        swap(ans[x], ans[b[i][0].second]);
        for (long long i = 0; i < n; ++i) {
          u[ans[i]] = i;
        }
        for (long long i = 0; i < n; ++i) {
          cout << u[i] + 1 << " ";
        }
        return 0;
      }
    }
  }
  cout << "YES" << endl;
  for (long long i = 0; i < n; ++i) {
    cout << u[i] + 1 << " ";
  }
  return 0;
}