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 = 200010;
set<pair<int, int> > mp;
vector<pair<int, int> > vec[N];
int a[N], b[N];
int st[30][N];
int ans[N];
int where[N];
int lg2[N];
int main() {
  int n;
  cin >> n;
  lg2[1] = 0;
  for (int i = 2; i <= n; i++) lg2[i] = lg2[i >> 1] + 1;
  for (int i = 1; i <= n; i++) {
    cin >> a[i] >> b[i];
    vec[a[i]].push_back({b[i], i});
  }
  set<pair<int, int> >::iterator it;
  for (int i = 1; i <= n; i++) {
    mp.insert(vec[i].begin(), vec[i].end());
    it = mp.begin();
    ans[it->second] = i;
    where[i] = it->second;
    mp.erase(mp.begin());
  }
  for (int i = 1; i <= n; i++) st[0][i] = a[where[i]];
  for (int i = 1; i <= 29; i++)
    for (int j = 1; j + (1 << i) - 1 <= n; j++)
      st[i][j] = min(st[i - 1][j], st[i - 1][j + (1 << (i - 1))]);
  for (int i = 1; i <= n; i++) {
    int l = i + 1, r = b[where[i]];
    if (l > r) continue;
    int len = r - l + 1;
    len = lg2[len];
    int minv = min(st[len][l], st[len][r - (1 << len) + 1]);
    if (minv <= i) {
      cout << "NO" << endl;
      for (int i = 1; i <= n; i++) cout << ans[i] << ' ';
      cout << endl;
      for (int k = l; k <= r; k++) {
        if (a[where[k]] <= i) {
          swap(ans[where[k]], ans[where[i]]);
          break;
        }
      }
      for (int i = 1; i <= n; i++) cout << ans[i] << ' ';
      cout << endl;
      return 0;
    }
  }
  cout << "YES" << endl;
  for (int i = 1; i <= n; i++) cout << ans[i] << ' ';
  cout << endl;
  return 0;
}