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;
const int N = 2e5 + 10, INF = 1e9 + 10;
int n;
int a[N], b[N];
int ans[N], pos[N];
pair<int, int> tree[N << 2];
vector<pair<int, int> > vec[N];
void build(int node, int l, int r) {
  if (l == r) {
    tree[node].first = a[pos[l]];
    tree[node].second = l;
    return;
  }
  int mid = l + r >> 1;
  build(node << 1, l, mid);
  build(node << 1 | 1, mid + 1, r);
  tree[node] = min(tree[node << 1], tree[node << 1 | 1]);
}
pair<int, int> query(int node, int l, int r, int L, int R) {
  if (L <= l && R >= r) return tree[node];
  int mid = l + r >> 1;
  pair<int, int> res1 = {INF, INF};
  pair<int, int> res2 = {INF, INF};
  if (L <= mid) res1 = query(node << 1, l, mid, L, R);
  if (R > mid) res2 = query(node << 1 | 1, mid + 1, r, L, R);
  return min(res1, res2);
}
void print() {
  for (int i = 1; i <= n; i++) {
    printf("%d ", ans[i]);
  }
  puts("");
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    scanf("%d%d", &a[i], &b[i]);
    vec[a[i]].push_back({b[i], i});
  }
  set<pair<int, int> > s;
  for (int i = 1; i <= n; i++) {
    s.insert(vec[i].begin(), vec[i].end());
    ans[s.begin()->second] = i;
    pos[i] = s.begin()->second;
    s.erase(s.begin());
  }
  build(1, 1, n);
  for (int i = 1; i <= n; i++) {
    pair<int, int> tmp = query(1, 1, n, i + 1, b[pos[i]]);
    if (tmp.first <= i) {
      puts("NO");
      print();
      swap(ans[pos[i]], ans[pos[tmp.second]]);
      print();
      return 0;
    }
  }
  puts("YES");
  print();
  return 0;
}