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;
int n, x[200005], y[200005], z[200005], la = -1, lb;
vector<pair<int, int>> v[200005];
priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>>
    q;
stack<pair<int, int>> s;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d%d", y + i, z + i);
    v[y[i]].emplace_back(z[i], i);
  }
  for (int i = 1; i <= n; i++) {
    for (auto& j : v[i]) q.push(j);
    int tmp = q.top().second;
    q.pop();
    while (!s.empty() && z[s.top().second] < i) s.pop();
    if (!s.empty() && y[tmp] <= s.top().first) {
      la = s.top().second;
      lb = tmp;
    }
    s.emplace(i, tmp);
    x[tmp] = i;
  }
  puts(la == -1 ? "YES" : "NO");
  for (int i = 1; i <= n; i++) printf("%d%c", x[i], " \n"[i == n]);
  if (la != -1)
    for (int i = 1; i <= n; i++)
      printf("%d%c", i == la ? x[lb] : i == lb ? x[la] : x[i], " \n"[i == n]);
}
