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;
int n;
vector<pair<int, int> > str[200005];
int a[200005];
int b[200005];
int ans[200005];
int p[200005];
pair<int, int> seg[400005];
int inf = 1e9;
void out() {
  for (int i = 1; i <= n; i++) cout << ans[i] << " ";
  cout << "\n";
}
pair<int, int> q(int l, int r) {
  l--;
  pair<int, int> res{inf, inf};
  for (l += n, r += n; l < r; l /= 2, r /= 2) {
    if (l % 2 == 1) res = min(res, seg[l++]);
    if (r % 2 == 1) res = min(res, seg[--r]);
  }
  return res;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cerr.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i] >> b[i];
    str[a[i]].push_back({b[i], i});
  }
  set<pair<int, int> > ac;
  for (int i = 1; i <= n; i++) {
    ac.insert(str[i].begin(), str[i].end());
    ans[ac.begin()->second] = i;
    p[i] = ac.begin()->second;
    ac.erase(ac.begin());
  }
  for (int i = 0; i < n; i++) {
    seg[i + n] = {a[p[i + 1]], i + 1};
  }
  for (int i = n - 1; i > 0; i--) {
    seg[i] = min(seg[2 * i], seg[2 * i + 1]);
  }
  for (int i = 1; i <= n; i++) {
    pair<int, int> j = q(i + 1, b[p[i]]);
    if (j.first == inf) continue;
    if (j.first <= i) {
      cout << "NO\n";
      out();
      swap(ans[p[i]], ans[p[j.second]]);
      out();
      return 0;
    }
  }
  cout << "YES\n";
  out();
  return 0;
}