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;
vector<pair<int, int> > vec[200005];
int pos[200005];
int ans1[200005], ans2[200005];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int a, b;
    scanf("%d %d", &a, &b);
    vec[a].push_back({b, i});
  }
  set<pair<int, int> > se;
  for (int i = 1; i <= n; i++) {
    for (pair<int, int> p : vec[i]) se.insert(p);
    pos[i] = se.begin()->second;
    se.erase(se.begin());
  }
  for (int i = 1; i <= n; i++) ans1[pos[i]] = i;
  for (int i = 1; i <= n; i++) ans2[i] = ans1[i];
  bool flag = true;
  for (int i = 1; i <= n; i++) {
    for (pair<int, int> p : vec[i]) se.insert(p);
    if (se.size() == 1) {
      se.erase(se.begin());
      continue;
    }
    auto it = se.begin();
    ++it;
    int j = it->second;
    if (ans1[j] <= se.begin()->first) {
      swap(ans2[pos[i]], ans2[j]);
      flag = false;
      break;
    }
    se.erase(se.begin());
  }
  printf("%s\n", flag ? "YES" : "NO");
  for (int i = 1; i <= n; i++) printf("%d ", ans1[i]);
  printf("\n");
  if (!flag) {
    for (int i = 1; i <= n; i++) printf("%d ", ans2[i]);
    printf("\n");
  }
  return 0;
}
