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>
struct Seg {
  int a;
  int b;
  int id;
};
int n;
const int maxN(200010);
Seg seg[maxN];
int order[maxN];
int p[maxN];
std::pair<int, int> tree[maxN * 2];
int ans[maxN];
bool cmpA(int a, int &b) { return seg[a].a < seg[b].a; }
bool operator>(const Seg &a, const Seg &b) { return a.b > b.b; }
void output() {
  for (int i(1); i <= n; ++i) ans[p[i]] = i;
  for (int i(1); i <= n; ++i) printf("%d ", ans[i]);
  printf("\n");
}
int query(int l, int r) {
  std::pair<int, int> ans{std::numeric_limits<int>::max(), -1};
  --l;
  for (l += n, r += n; l < r; l >>= 1, r >>= 1) {
    if (l & 1) ans = std::min(ans, tree[l++]);
    if (r & 1) ans = std::min(ans, tree[--r]);
  }
  return ans.second;
}
int main() {
  scanf("%d", &n);
  for (int i(1); i <= n; ++i) {
    scanf("%d %d", &seg[i].a, &seg[i].b);
    seg[i].id = i;
    order[i] = i;
  }
  std::sort(order + 1, order + n + 1, cmpA);
  std::priority_queue<Seg, std::vector<Seg>, std::greater<Seg> > heap;
  for (int i(1), j(1); i <= n; ++i) {
    while (j <= n && seg[order[j]].a <= i) {
      heap.push(seg[order[j]]);
      ++j;
    }
    tree[n + i - 1].first = heap.top().a;
    tree[n + i - 1].second = i;
    p[i] = heap.top().id;
    heap.pop();
  }
  for (int i(n - 1); i >= 1; --i) {
    tree[i] = std::min(tree[i << 1], tree[i << 1 | 1]);
  }
  for (int i(1); i <= n; ++i) {
    int j(query(i + 1, seg[p[i]].b));
    if (j != -1 && seg[p[j]].a <= i) {
      printf("NO\n");
      output();
      std::swap(p[i], p[j]);
      output();
      return 0;
    }
  }
  printf("YES\n");
  output();
  return 0;
}