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 N = 2e5 + 7, mod = 1e9 + 7;
int n, l[N], r[N], ans[N], nxt[N];
vector<pair<int, int> > p[N];
priority_queue<pair<int, int>, vector<pair<int, int> >,
               greater<pair<int, int> > >
    pr;
void print() {
  for (int i = 1; i <= n; i++) {
    cout << ans[i] << " ";
  }
  cout << "\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> l[i] >> r[i];
    p[l[i]].push_back(make_pair(r[i], i));
  }
  for (int i = 1; i <= n; i++) {
    for (auto t : p[i]) {
      pr.push(t);
    }
    int now = pr.top().second;
    ans[now] = i;
    pr.pop();
    if (!pr.empty()) {
      nxt[now] = pr.top().second;
    }
  }
  for (int i = 1; i <= n; i++) {
    if (nxt[i] == 0) continue;
    int nx = nxt[i];
    if (l[nx] <= ans[i] && r[nx] >= ans[i] && l[i] <= ans[nx] &&
        r[i] >= ans[nx]) {
      cout << "NO\n";
      print();
      swap(ans[i], ans[nx]);
      print();
      return 0;
    }
  }
  cout << "YES\n";
  print();
}
