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;
pair<int, int> label[200005];
pair<pair<int, int>, int> sortedLabel[200005];
int order[200005];
int lastInd[200005];
int ans[200005];
int n;
int tree[400005];
int query(int l, int r) {
  int res = 0;
  for (l += n, r += n; l < r; l >>= 1, r >>= 1) {
    if (l & 1) res = max(res, tree[l++]);
    if (r & 1) res = max(res, tree[--r]);
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> label[i].first >> label[i].second;
    label[i].first--;
    label[i].second--;
    sortedLabel[i].first = label[i];
    sortedLabel[i].second = i;
  }
  sort(sortedLabel, sortedLabel + n);
  set<pair<pair<int, int>, int> > end_start;
  int cur = 0;
  for (int i = 0; i < n; i++) {
    while (cur < n && sortedLabel[cur].first.first <= i) {
      end_start.insert(
          {{sortedLabel[cur].first.second, sortedLabel[cur].first.first},
           sortedLabel[cur].second});
      cur++;
    }
    auto x = end_start.begin();
    order[i] = x->second;
    end_start.erase(x);
  }
  for (int i = 0; i < n; i++) {
    lastInd[i] = label[order[i]].second;
  }
  for (int i = 0; i < n; i++) tree[n + i] = lastInd[i];
  for (int i = n - 1; i > 0; i--) tree[i] = max(tree[i << 1], tree[i << 1 | 1]);
  pair<int, int> xchng = {-1, -1};
  for (int i = 1; i < n; i++) {
    int maxVal = query(label[order[i]].first, i);
    if (maxVal >= i) {
      xchng.first = i;
      for (int j = label[order[i]].first; j < i; j++) {
        if (label[order[j]].second >= i) {
          xchng.second = j;
          break;
        }
      }
      break;
    }
  }
  for (int i = 0; i < n; i++) ans[order[i]] = i;
  if (xchng.second != -1) {
    cout << "NO\n";
    for (int i = 0; i < n; i++) cout << ans[i] + 1 << " ";
    cout << "\n";
    swap(order[xchng.first], order[xchng.second]);
    for (int i = 0; i < n; i++) ans[order[i]] = i;
    for (int i = 0; i < n; i++) cout << ans[i] + 1 << " ";
  } else {
    cout << "YES\n";
    for (int i = 0; i < n; i++) cout << ans[i] + 1 << " ";
  }
  return 0;
}