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;
using ll = long long;
using ii = pair<int, int>;
const int N = 2e5 + 10;
int n, l[N], r[N], p[N];
vector<ii> interval[N];
multiset<ii> st;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d %d", &l[i], &r[i]);
    interval[l[i]].push_back({r[i], i});
  }
  for (int i = 1; i <= n; i++) {
    while (!interval[i].empty()) {
      st.insert(interval[i].back());
      interval[i].pop_back();
    }
    auto it = *st.begin();
    st.erase(st.begin());
    p[it.second] = i;
  }
  for (int i = 1; i <= n; i++) {
    interval[p[i]].push_back({i, +1});
    interval[r[i] + 1].push_back({i, -1});
  }
  vector<int> ord(n, 0);
  iota(ord.begin(), ord.end(), 1);
  sort(ord.begin(), ord.end(), [&](int x, int y) { return p[x] < p[y]; });
  st.clear();
  int p1 = -1, p2;
  for (int i : ord) {
    for (auto work : interval[p[i]])
      if (work.second == -1) st.erase(st.find({p[work.first], work.first}));
    if (!st.empty()) {
      auto it = *st.rbegin();
      int mx = it.first;
      if (l[i] <= mx) {
        p1 = i;
        p2 = it.second;
        break;
      }
    }
    for (auto work : interval[p[i]])
      if (work.second == 1) st.insert({p[work.first], work.first});
  }
  if (p1 + 1) {
    puts("NO");
    for (int i = 1; i <= n; i++) printf("%d ", p[i]);
    puts("");
    swap(p[p1], p[p2]);
    for (int i = 1; i <= n; i++) printf("%d ", p[i]);
    puts("");
  } else {
    puts("YES");
    for (int i = 1; i <= n; i++) printf("%d ", p[i]);
    puts("");
  }
  return 0;
}