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 
#include <bits/stdc++.h>
using namespace std;
const long long mod = 998244353;
int solve();
void precomp();
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int t = 1;
  if (0) cin >> t;
  precomp();
  while (t--) solve();
}
void precomp() { return; }
int n;
vector<pair<int, int> > v[200005];
int l[200005], r[200005], ans[200005];
priority_queue<pair<int, int> > pq;
vector<pair<int, int> > swp;
int solve() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    int a, b;
    cin >> a >> b;
    l[i] = a, r[i] = b;
    v[a].push_back({b, i});
  }
  for (int i = 1; i <= n; i++) {
    for (pair<int, int> j : v[i]) pq.push({-j.first, j.second});
    int nw = pq.top().second;
    pq.pop();
    ans[nw] = i;
    if (!pq.empty()) swp.push_back({nw, pq.top().second});
  }
  for (pair<int, int> i : swp) {
    swap(ans[i.first], ans[i.second]);
    if (l[i.first] <= ans[i.first] && ans[i.first] <= r[i.first] &&
        l[i.second] <= ans[i.second] && ans[i.second] <= r[i.second]) {
      cout << "NO\n";
      for (int j = 0; j < n; j++) cout << ans[j] << " ";
      cout << endl;
      swap(ans[i.first], ans[i.second]);
      for (int j = 0; j < n; j++) cout << ans[j] << " ";
      cout << endl;
      return 0;
    }
    swap(ans[i.first], ans[i.second]);
  }
  cout << "YES\n";
  for (int i = 0; i < n; i++) cout << ans[i] << " ";
  cout << endl;
  return 0;
}
