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;
int unx() {
  int x = 0, n = 0;
  char c = getchar();
  for (; !isdigit(c); c = getchar()) n = (c == '-');
  for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
  return (n) ? -x : x;
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int rnd(int l, int r) { return l + rng() % (r - l + 1); }
const int N = 2e5 + 5;
int n;
pair<int, int> a[N];
int s[N], pos[N], id[N];
priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>>
    h;
vector<pair<int, int>> event[N];
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  for (int i = 1, _c = n = unx(); i <= _c; ++i) {
    a[i] = {unx(), unx()};
  }
  for (int i = 1, _c = n; i <= _c; ++i)
    event[a[i].first].push_back(make_pair(a[i].second, i));
  for (int i = 1, _c = n; i <= _c; ++i) {
    for (auto &v : event[i]) h.push(v);
    int ri, index;
    tie(ri, index) = h.top();
    h.pop();
    id[index] = i;
    pos[i] = index;
    if (h.size()) s[index] = h.top().second;
  }
  for (int i = 1, _c = n; i <= _c; ++i) {
    int cur = i, nxt = s[i];
    if (nxt && a[cur].first <= id[nxt] && id[nxt] <= a[cur].second &&
        a[nxt].first <= id[cur] && id[cur] <= a[nxt].second) {
      cout << "NO\n";
      for (int i = 1, _c = n; i <= _c; ++i) cout << id[i] << ' ';
      cout << '\n';
      swap(id[cur], id[nxt]);
      for (int i = 1, _c = n; i <= _c; ++i) cout << id[i] << ' ';
      return 0;
    }
  }
  cout << "YES\n";
  for (int i = 1, _c = n; i <= _c; ++i) cout << id[i] << ' ';
}