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 N = 2e5 + 9;
int n, ans1 = -1, ans2, par[N];
pair<int, int> m, p[N], tree[N * 2];
vector<pair<int, int> > v[N];
set<pair<int, int> > s;
pair<int, int> query(int l, int r) {
  pair<int, int> res;
  res.first = res.second = -1;
  r++;
  for (l += n, r += n; l < r; l >>= 1, r >>= 1) {
    if (l & 1) {
      if (tree[l].first > res.first) res = tree[l];
      l++;
    }
    if (r & 1) {
      if (tree[--r].first > res.first) res = tree[r];
    }
  }
  return res;
}
void make_tree() {
  for (int i = n - 1; i >= 1; i--) {
    pair<int, int> p1 = tree[i * 2], p2 = tree[i * 2 + 1];
    if (p1.first > p2.first)
      tree[i] = p1;
    else
      tree[i] = p2;
  }
}
int main() {
  cin >> n;
  for (long long i = 0; i < n; i++) {
    scanf("%d%d", &p[i].first, &p[i].second);
    ;
    p[i].first--, p[i].second--;
    v[p[i].first].push_back(make_pair(p[i].second, i));
  }
  for (long long i = 0; i < n; i++) {
    for (long long j = 0; j < v[i].size(); j++) s.insert(v[i][j]);
    pair<int, int> u = *s.begin();
    s.erase(u);
    tree[i + n] = u;
    par[u.second] = i;
  }
  make_tree();
  for (long long i = 0; i < n; i++) {
    m = query(p[i].first, par[i] - 1);
    if (m.first >= par[i]) ans1 = i, ans2 = m.second;
  }
  if (ans1 == -1)
    cout << "YES" << endl;
  else
    cout << "NO" << endl;
  for (long long i = 0; i < n; i++) cout << par[i] + 1 << " ";
  cout << endl;
  if (ans1 != -1) {
    swap(par[ans1], par[ans2]);
    for (long long i = 0; i < n; i++) cout << par[i] + 1 << " ";
  }
}