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;
void print_ans(const vector<int>& p) {
  vector<int> ans(p.size());
  for (int i = 0; i < (int)p.size(); ++i) ans[p[i]] = i;
  for (int i = 0; i < (int)ans.size(); ++i)
    printf("%d%c", ans[i] + 1, i + 1 == (int)ans.size() ? '\n' : ' ');
}
int main() {
  int n;
  scanf("%d", &n);
  vector<int> a(n), b(n);
  vector<pair<int, int>> events;
  for (int i = 0; i < n; ++i) {
    scanf("%d %d", &a[i], &b[i]);
    --a[i], --b[i];
    events.emplace_back(a[i], i);
  }
  sort(events.begin(), events.end());
  vector<int> p;
  priority_queue<pair<int, int>> pq;
  for (int i = 0, e = 0; i < n; ++i) {
    while (e < (int)events.size() && events[e].first <= i) {
      int v = events[e].second;
      pq.emplace(-b[v], v);
      ++e;
    }
    int v = pq.top().second;
    pq.pop();
    p.emplace_back(v);
  }
  int lef = -1, rig = -1;
  stack<int> vrig;
  for (int i = n - 1; i >= 0; --i) {
    int v = p[i];
    while (!vrig.empty() && a[p[vrig.top()]] > i) vrig.pop();
    if (!vrig.empty() && vrig.top() <= b[v]) {
      lef = i, rig = vrig.top();
      break;
    }
    while (!vrig.empty() && a[p[vrig.top()]] >= a[v]) vrig.pop();
    vrig.push(i);
  }
  if (lef == -1) {
    puts("YES");
  } else {
    puts("NO");
    print_ans(p);
    swap(p[lef], p[rig]);
  }
  print_ans(p);
  return 0;
}
