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 
#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 100;
const int inf = 1e9 + 7;
int n, a[N], b[N], pos[N], ans[N];
vector<pair<int, int> > bs[N];
pair<int, int> seg[2 * N];
int query(int l, int r) {
  l += n - 1;
  r += n;
  pair<int, int> res = {inf, inf};
  for (; l < r; l >>= 1, r >>= 1) {
    if (l & 1) res = min(res, seg[l++]);
    if (r & 1) res = min(res, seg[--r]);
  }
  return res.second;
}
void show() {
  for (int i = 1; i <= n; i++) ans[pos[i]] = i;
  for (int i = 1; i <= n; i++) cout << ans[i] << ' ';
  cout << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i] >> b[i];
    bs[a[i]].push_back({b[i], i});
  }
  set<pair<int, int> > st;
  for (int i = 1; i <= n; i++) {
    st.insert(bs[i].begin(), bs[i].end());
    auto bg = *st.begin();
    pos[i] = bg.second;
    st.erase(bg);
  }
  for (int i = 0; i < n; i++) {
    seg[i + n] = {a[pos[i + 1]], i + 1};
  }
  for (int i = n - 1; i > 0; i--) {
    seg[i] = min(seg[i << 1], seg[i << 1 | 1]);
  }
  for (int i = 1; i <= n; i++) {
    int id = query(i + 1, b[pos[i]]);
    if (id != inf && a[pos[id]] <= i) {
      cout << "NO" << endl;
      show();
      swap(pos[i], pos[id]);
      show();
      exit(0);
    }
  }
  cout << "YES" << endl;
  show();
}
