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> st[1000005], mn;
pair<pair<int, int>, int> a[200005];
pair<int, pair<pair<int, int>, int> > b[200005];
int n, kq[200005];
set<int> s;
void update(int id, int l, int r, int vt, pair<int, int> gt) {
  if (l > r || l > vt || r < vt) return;
  if (l == r) {
    st[id] = min(st[id], gt);
    return;
  }
  int mid = (l + r) / 2;
  update(id * 2, l, mid, vt, gt);
  update(id * 2 + 1, mid + 1, r, vt, gt);
  st[id] = min(st[id * 2], st[id * 2 + 1]);
}
pair<int, int> get(int id, int l, int r, int u, int v) {
  if (l > r || l > v || r < u) return {1e9, 0};
  if (l >= u && r <= v) return st[id];
  int mid = (l + r) / 2;
  return min(get(id * 2, l, mid, u, v), get(id * 2 + 1, mid + 1, r, u, v));
}
bool cmp(pair<pair<int, int>, int> x, pair<pair<int, int>, int> y) {
  if (x.first.second != y.first.second) return x.first.second < y.first.second;
  return x.first.first < y.first.first;
}
void build(int id, int l, int r) {
  if (l > r) return;
  st[id] = {1e9, 0};
  if (l == r) return;
  int mid = (l + r) / 2;
  build(id * 2, l, mid);
  build(id * 2 + 1, mid + 1, r);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i].first.first >> a[i].first.second;
    a[i].second = i;
  }
  sort(a + 1, a + n + 1, cmp);
  for (int i = 1; i <= n; ++i) s.insert(i);
  for (int i = 1; i <= n; ++i) {
    kq[a[i].second] = (*s.lower_bound(a[i].first.first));
    s.erase(kq[a[i].second]);
    b[i] = {kq[a[i].second], a[i]};
  }
  build(1, 1, n);
  sort(b + 1, b + n + 1);
  for (int i = n; i > 0; --i) {
    mn = get(1, 1, n, b[i].first + 1, b[i].second.first.second);
    if (mn.first <= b[i].first) {
      cout << "NO" << '\n';
      for (int i = 1; i <= n; ++i) cout << kq[i] << ' ';
      cout << '\n';
      swap(kq[b[i].second.second], kq[b[mn.second].second.second]);
      for (int i = 1; i <= n; ++i) cout << kq[i] << ' ';
      return 0;
    }
    update(1, 1, n, b[i].first, {b[i].second.first.first, i});
  }
  cout << "YES" << '\n';
  for (int i = 1; i <= n; ++i) cout << kq[i] << ' ';
  return 0;
}