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;
int n, kezd[200002], veg[200002], pos[200002], inv[200002], x, csa, csb;
vector<pair<pair<int, int>, int> > sz;
vector<int> v;
priority_queue<pair<int, int> > q;
void p() {
  for (int i = 1; i <= n; i++) inv[pos[i]] = i;
  for (int i = 1; i <= n; i++) cout << inv[i] << " ";
  cout << "\n";
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int a, b;
    cin >> a >> b, kezd[i] = a, veg[i] = b;
    sz.push_back({{a, b}, i});
  }
  sort(sz.begin(), sz.end());
  for (int i = 1; i <= n; i++) {
    while (x < n && sz[x].first.first == i)
      q.push({-sz[x].first.second, sz[x].second}), x++;
    pos[i] = q.top().second, q.pop();
  }
  for (int i = 1; i <= n; i++) {
    while (v.size() > 0 && veg[pos[v.back()]] < i) {
      v.pop_back();
    }
    int a = kezd[pos[i]], b = veg[pos[i]];
    if (v.size() > 0 && a <= v.back()) {
      csa = i, csb = v.back();
    }
    while (v.size() > 0 && b >= veg[pos[v.back()]]) v.pop_back();
    v.push_back(i);
  }
  if (!csa) {
    cout << "YES"
         << "\n";
    p();
  } else {
    cout << "NO"
         << "\n";
    p(), swap(pos[csa], pos[csb]), p();
  }
  return 0;
}
