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 INF = 1e9;
const long long LINF = 1e18;
int n;
int A[200003], B[200003];
int ans1[200003];
int ans2[200003];
vector<pair<int, int> > op[200003];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d %d", &A[i], &B[i]);
    op[A[i]].push_back({B[i], i});
  }
  vector<pair<int, int> > vec;
  bool bol = 1;
  set<pair<int, int> > st;
  for (int i = 1; i <= n; i++) {
    for (pair<int, int> x : op[i]) st.insert(x);
    while (vec.size() && B[vec.back().second] < i) {
      ans1[vec.back().second] = vec.back().first;
      ans2[vec.back().second] = vec.back().first;
      vec.pop_back();
    }
    if (vec.size() && A[st.begin()->second] <= vec.back().first) {
      pair<int, int> x = *st.begin();
      bol = 0;
      ans1[vec.back().second] = vec.back().first;
      ans2[vec.back().second] = i;
      ans2[x.second] = vec.back().first;
      ans1[x.second] = i;
      st.erase(x);
      vec.pop_back();
    } else if (st.size() == 1 || st.begin()->first == i) {
      pair<int, int> x = *st.begin();
      ans1[x.second] = i;
      ans2[x.second] = i;
      st.erase(x);
    } else {
      pair<int, int> x = *st.begin();
      vec.push_back({i, x.second});
      st.erase(x);
    }
  }
  while (vec.size()) {
    ans1[vec.back().second] = vec.back().first;
    ans2[vec.back().second] = vec.back().first;
    vec.pop_back();
  }
  if (bol) {
    printf("YES\n");
    for (int i = 1; i <= n; i++) printf("%d%c", ans1[i], (i == n) ? '\n' : ' ');
  } else {
    printf("NO\n");
    for (int i = 1; i <= n; i++) printf("%d%c", ans1[i], (i == n) ? '\n' : ' ');
    for (int i = 1; i <= n; i++) printf("%d%c", ans2[i], (i == n) ? '\n' : ' ');
  }
}