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;
vector<pair<int, int> > v[200003];
int n, ans[200003], a[200003], b[200003], res[262144];
pair<int, int> dat[524288];
pair<int, int> query(int x, int y, int a, int b, int k) {
  if (x <= a && b <= y) return dat[k];
  if (a > y || b < x) return make_pair(0, 233);
  return max(query(x, y, a, (a + b) / 2, k * 2),
             query(x, y, (a + b) / 2 + 1, b, k * 2 + 1));
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    scanf("%d%d", a + i, b + i);
    a[i]--;
    b[i]--;
    v[b[i]].push_back(make_pair(a[i], i));
  }
  priority_queue<pair<int, int> > q;
  for (int i = n - 1; i >= 0; i--) {
    for (int j = 0; j < v[i].size(); j++) q.push(v[i][j]);
    ans[q.top().second] = i;
    q.pop();
  }
  memset(res, -1, sizeof(res));
  for (int i = 0; i < n; i++) res[ans[i]] = i;
  for (int i = 0; i < n; i++) dat[i + 262144] = make_pair(b[res[i]], res[i]);
  for (int i = 262143; i >= 1; i--) dat[i] = max(dat[i * 2], dat[i * 2 + 1]);
  for (int i = 1; i < n; i++) {
    if (a[res[i]] == i) continue;
    pair<int, int> maxn = query(a[res[i]], i - 1, 0, 262143, 1);
    if (maxn.first >= i) {
      cout << "NO\n";
      for (int j = 0; j < n; j++) printf("%d ", ans[j] + 1);
      cout << endl;
      swap(ans[res[i]], ans[maxn.second]);
      for (int j = 0; j < n; j++) printf("%d ", ans[j] + 1);
      return 0;
    }
  }
  cout << "YES\n";
  for (int j = 0; j < n; j++) printf("%d ", ans[j] + 1);
  cout << "\n";
}