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> T[4 * 200005];
void update(int n, int b, int e, int p, int v) {
  if (b == p && e == p) {
    T[n] = make_pair(v, p);
  } else {
    int m = (b + e) / 2;
    if (p <= m)
      update(2 * n, b, m, p, v);
    else
      update(2 * n + 1, m + 1, e, p, v);
    T[n] = max(T[2 * n], T[2 * n + 1]);
  }
}
pair<int, int> query(int n, int b, int e, int l, int r) {
  if (b == l && e == r) {
    return T[n];
  } else {
    int m = (b + e) / 2;
    if (m >= r) return query(2 * n, b, m, l, r);
    if (l > m) return (query(2 * n + 1, m + 1, e, l, r));
    return max(query(2 * n, b, m, l, m), query(2 * n + 1, m + 1, e, m + 1, r));
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  vector<pair<pair<int, int>, int> > arr(n);
  vector<pair<int, int> > ori(n);
  int a, b;
  for (int i = 0; i < n; i++) {
    cin >> a >> b;
    arr[i] = make_pair(make_pair(a, b), i);
    ori[i] = make_pair(a, b);
  }
  sort(arr.begin(), arr.end());
  priority_queue<pair<int, int> > q;
  vector<int> result1(n);
  vector<int> result2(n);
  int index = 0;
  map<int, int> comp;
  for (int i = 1; i <= n; i++) {
    while (index < n && arr[index].first.first == i) {
      q.push(make_pair(-1 * arr[index].first.second, arr[index].second));
      index++;
    }
    pair<int, int> current = q.top();
    q.pop();
    result1[current.second] = i;
    comp[i] = current.second;
  }
  for (int i = 0; i < n; i++) {
    result2[i] = result1[i];
  }
  bool one = true;
  for (int i = 1; i <= n; i++) {
    int l = ori[comp[i]].first;
    int r = ori[comp[i]].second;
    pair<int, int> res = query(1, 0, 200005, l, r);
    if (res.first >= i) {
      one = false;
      result2[comp[i]] = result1[comp[res.second]];
      result2[comp[res.second]] = result1[comp[i]];
      break;
    }
    update(1, 0, 200005, i, r);
  }
  if (one) {
    cout << "YES" << '\n';
    for (int i = 0; i < n; i++) {
      cout << result1[i] << " ";
    }
    cout << '\n';
  } else {
    cout << "NO" << '\n';
    for (int i = 0; i < n; i++) {
      cout << result1[i] << " ";
    }
    cout << '\n';
    for (int i = 0; i < n; i++) {
      cout << result2[i] << " ";
    }
    cout << '\n';
  }
  return 0;
}