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 INF = 0x3f3f3f3f, N = 2e5 + 7;
int l[N], r[N];
int bit[N];
void add(int pos, int val) {
  for (; pos < N; pos += pos & -pos) bit[pos] = min(bit[pos], val);
}
int qry(int pos) {
  int mn = INF;
  for (; pos; pos -= pos & -pos) mn = min(mn, bit[pos]);
  return mn;
}
signed main() {
  std::ios::sync_with_stdio(0);
  std::cin.tie(0);
  ;
  int n;
  cin >> n;
  vector<array<int, 3>> v;
  for (int i = 0; i < n; i++) {
    cin >> l[i] >> r[i];
    v.push_back({l[i], r[i], i});
  }
  sort((v).begin(), (v).end());
  priority_queue<pair<int, int>, vector<pair<int, int>>,
                 greater<pair<int, int>>>
      pq;
  vector<int> p;
  for (int i = 1, j = 0; i <= n; i++) {
    while (j < v.size() && v[j][0] <= i) {
      pq.push({v[j][1], v[j][2]});
      j++;
    }
    assert(pq.size());
    pair<int, int> pp = pq.top();
    pq.pop();
    p.push_back(pp.second);
  }
  auto print = [&]() {
    vector<int> ans(n);
    for (int i = 0; i < n; i++) ans[p[i]] = i + 1;
    for (int i : ans) cout << i << " ";
    cout << endl;
  };
  memset(bit, 0x3f, sizeof bit);
  for (int i = n - 1; i >= 0; i--) {
    int qq = qry(i + 1);
    if (qq <= r[p[i]]) {
      cout << "NO" << endl;
      print();
      swap(p[i], p[qq - 1]);
      print();
      return 0;
    }
    add(l[p[i]], i + 1);
  }
  cout << "YES" << endl;
  print();
}
