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;
constexpr static int MAXN = 2e5 + 10;
int n;
int a[MAXN], b[MAXN];
vector<pair<int, int>> cur[MAXN];
int pos[MAXN];
int r[MAXN];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i] >> b[i];
    cur[a[i]].push_back({b[i], i});
  }
  for (int i = 0; i < n; i++) pos[i] = -1;
  int s1 = -1, s2 = -1;
  priority_queue<pair<int, int>> pq;
  for (int i = 1; i <= n; i++) {
    for (auto p : cur[i]) pq.push({-p.first, p.second});
    assert(pq.size() > 0);
    auto t = pq.top();
    pq.pop();
    r[t.second] = i;
    if (s1 != -1) continue;
    if (pq.size() != 0 && pq.top().first == t.first) {
      s1 = t.second;
      s2 = pq.top().second;
      continue;
    }
    int prv = pos[t.second];
    if (prv != -1 && i <= b[prv] && i >= a[prv]) {
      s1 = prv;
      s2 = t.second;
      continue;
    }
    if (pq.size() == 0 || -t.first == i) continue;
    pos[pq.top().second] = t.second;
  }
  cout << (s1 == -1 ? "YES" : "NO") << endl;
  for (int i = 0; i < n; i++) cout << r[i] << ' ';
  cout << endl;
  if (s1 != -1) {
    swap(r[s1], r[s2]);
    for (int i = 0; i < n; i++) cout << r[i] << ' ';
    cout << endl;
  }
}