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 inf = 0x3f3f3f3f;
const long long mod = 1000000007;
const long long linf = 0x3f3f3f3f3f3f3f3f;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int n;
int ans[200100];
int a[200100], b[200100];
set<pair<int, int> > s;
pair<int, int> e[200100];
pair<int, int> change;
set<pair<int, int> > l, r;
pair<int, int> pos[200100];
void solve() {
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i] >> b[i];
  for (int i = 0; i < n; i++) e[i].first = a[i], e[i].second = i;
  sort(e, e + n);
  int j = 1;
  for (int i = 0; i < n; i++) {
    while (e[i].first > j) {
      pair<int, int> r = *s.begin();
      s.erase(r);
      ans[r.second] = j;
      j++;
    }
    s.insert({b[e[i].second], e[i].second});
  }
  for (auto r : s) {
    ans[r.second] = j;
    j++;
  }
  for (int i = 0; i < n; i++) {
    pos[i].first = ans[i];
    pos[i].second = i;
  }
  sort(pos, pos + n);
  change = {-1, -1};
  for (int i = 0; i < n; i++) {
    while (!r.empty() && pos[i].first > (*r.begin()).first) {
      int j = (*r.begin()).second;
      l.erase({pos[j].first, j});
      r.erase(r.begin());
    }
    if (!l.empty()) {
      auto it = l.lower_bound({a[pos[i].second], 0});
      if (it != l.end()) {
        change.first = pos[i].second;
        change.second = pos[(*it).second].second;
        break;
      }
    }
    l.insert({pos[i].first, i});
    r.insert({b[pos[i].second], i});
  }
  if (change.first != -1) {
    cout << "NO\n";
    for (int i = 0; i < n; i++) {
      cout << ans[i] << " ";
    }
    cout << "\n";
    swap(ans[change.first], ans[change.second]);
    for (int i = 0; i < n; i++) {
      cout << ans[i] << " ";
    }
    cout << "\n";
  } else {
    cout << "YES\n";
    for (int i = 0; i < n; i++) {
      cout << ans[i] << " ";
    }
    cout << "\n";
  }
}
int main() {
  cin.tie(0)->sync_with_stdio(0);
  ;
  solve();
  return 0;
}