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>
const int N = 2e5 + 10;
const int INF = 0x3f3f3f3f;
int n;
int st[N], en[N];
std::vector<std::pair<int, int> > v[N];
int ans[N], re_ans[N];
std::pair<int, int> minn[N << 1];
void Build() {
  for (int i = 1; i <= n; ++i)
    minn[i + n - 1] = std::make_pair(st[re_ans[i]], i);
  for (int i = n - 1; i >= 1; --i)
    minn[i] = std::min(minn[i << 1], minn[i << 1 | 1]);
}
int Find(int l, int r) {
  l += n, r += n;
  std::pair<int, int> res(INF, INF);
  for (; l < r; l >>= 1, r >>= 1) {
    if (l & 1) res = std::min(res, minn[l++]);
    if (r & 1) res = std::min(res, minn[--r]);
  }
  return res.second == INF ? -1 : res.second;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d%d", &st[i], &en[i]);
  for (int i = 1; i <= n; ++i) v[st[i]].push_back(std::make_pair(en[i], i));
  std::set<std::pair<int, int> > s;
  for (int i = 1; i <= n; ++i) {
    s.insert(v[i].begin(), v[i].end());
    ans[s.begin()->second] = i;
    re_ans[i] = s.begin()->second;
    s.erase(s.begin());
  }
  Build();
  for (int i = 1; i <= n; ++i) {
    int p = Find(i, en[re_ans[i]]);
    if (p == -1) continue;
    if (st[re_ans[p]] <= i) {
      printf("NO\n");
      for (int i = 1; i <= n; ++i) printf("%d ", ans[i]);
      printf("\n");
      std::swap(ans[re_ans[i]], ans[re_ans[p]]);
      for (int i = 1; i <= n; ++i) printf("%d ", ans[i]);
      printf("\n");
      return 0;
    }
  }
  printf("YES\n");
  for (int i = 1; i <= n; ++i) printf("%d ", ans[i]);
  printf("\n");
  return 0;
}