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;
int N;
int a[200005], b[200005];
int label[200005];
int label2[200005];
void perfectMatch() {
  deque<pair<pair<int, int>, int>> v;
  for (int i = 1; i <= N; i++) v.push_back({{a[i], b[i]}, i});
  sort(v.begin(), v.end());
  multiset<pair<int, int>> mst;
  for (int i = 1; i <= N; i++) {
    while ((int)v.size() > 0 && v[0].first.first <= i) {
      mst.insert({v[0].first.second, v[0].second});
      v.pop_front();
    }
    label2[i] = mst.begin()->second;
    label[mst.begin()->second] = i;
    mst.erase(mst.find(*mst.begin()));
  }
}
set<int> active;
int vis[200005];
int from[200005];
int ans[200005];
void foundAnother(int node, int nextNode) {
  vector<int> v;
  int cur = node;
  while (cur != label2[nextNode]) {
    v.push_back(cur);
    cur = from[cur];
  }
  reverse(v.begin(), v.end());
  v.push_back(label2[nextNode]);
  reverse(v.begin(), v.end());
  cout << "NO" << endl;
  for (int i = 1; i <= N; i++) cout << label[i] << ' ';
  cout << endl;
  for (int i = 1; i <= N; i++) ans[i] = label[i];
  int temp = ans[v.back()];
  for (int i = 0; i < (int)v.size(); i++) swap(temp, ans[v[i]]);
  for (int i = 1; i <= N; i++) cout << ans[i] << ' ';
  cout << endl;
  exit(0);
}
void dfs(int node) {
  vis[node] = 1;
  queue<int> toRemove;
  for (;;) {
    auto it = active.lower_bound(a[node]);
    if (it != active.end() && *it == label[node]) {
      toRemove.push(*it);
      it++;
    }
    if (it == active.end() || *it > b[node]) break;
    if (vis[label2[*it]] == 1) foundAnother(node, *it);
    toRemove.push(*it);
    if (vis[label2[*it]] == 2) continue;
    from[label2[*it]] = node;
    dfs(label2[*it]);
  }
  while (!toRemove.empty()) {
    if (active.count(toRemove.front())) active.erase(toRemove.front());
    toRemove.pop();
  }
  vis[node] = 2;
}
int main() {
  cin >> N;
  for (int i = 1; i <= N; i++) cin >> a[i] >> b[i];
  perfectMatch();
  for (int i = 1; i <= N; i++) active.insert(i);
  for (int i = 1; i <= N; i++)
    if (vis[i] == 0) dfs(i);
  cout << "YES" << endl;
  for (int i = 1; i <= N; i++) cout << label[i] << ' ';
  cout << endl;
}