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;
const int N = 2e5 + 5, inf = 1e9 + 7;
set<pair<int, int> > hllw;
vector<pair<int, int> > vec[200005];
int n, A[200005], B[200005];
int ans[200005], where[200005];
pair<int, int> st[400005];
int query(int l, int r) {
  l--;
  pair<int, int> res;
  res.first = inf;
  res.second = inf;
  for (l += n, r += n; l < r; l >>= 1, r >>= 1) {
    if (l & 1) res = min(res, st[l++]);
    if (r & 1) res = min(res, st[--r]);
  }
  return res.second;
}
void cout_ans() {
  for (int i = 1; i <= n; i++) cout << ans[i] << " ";
  cout << endl;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> A[i] >> B[i];
    vec[A[i]].push_back(make_pair(B[i], i));
  }
  cout << endl;
  for (int i = 1; i <= n; i++) {
    hllw.insert(vec[i].begin(), vec[i].end());
    ans[hllw.begin()->second] = i;
    where[i] = hllw.begin()->second;
    hllw.erase(hllw.begin());
  }
  for (int i = 0; i < n; i++) {
    st[i + n].first = A[where[i + 1]];
    st[i + n].second = i + 1;
  }
  for (int i = n - 1; i > 0; i--) {
    st[i] = min(st[i << 1 | 1], st[i << 1]);
  }
  for (int i = 1; i <= n; i++) {
    int j = query(i + 1, B[where[i]]);
    if (j == inf) continue;
    if (A[where[j]] <= i) {
      cout << "NO" << endl;
      cout_ans();
      swap(ans[where[i]], ans[where[j]]);
      cout_ans();
      return 0;
    }
  }
  cout << "YES" << endl;
  cout_ans();
  return 0;
}