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 inf = 0x3f3f3f3f;
const int maxn = 2e5 + 10;
int tree[maxn << 2], id[maxn << 2];
int ans[maxn], pos[maxn], L[maxn], R[maxn];
int n;
vector<pair<int, int> > v[maxn];
void build(int o, int l, int r) {
  if (l == r) {
    tree[o] = L[pos[l]];
    id[o] = l;
    return;
  }
  int mid = l + r >> 1;
  build(o << 1, l, mid);
  build(o << 1 | 1, mid + 1, r);
  if (tree[o << 1] < tree[o << 1 | 1])
    id[o] = id[o << 1];
  else
    id[o] = id[o << 1 | 1];
  tree[o] = min(tree[o << 1 | 1], tree[o << 1]);
}
pair<int, int> query(int o, int l, int r, int x, int y) {
  if (x <= l && y >= r) return make_pair(tree[o], id[o]);
  int mid = l + r >> 1;
  pair<int, int> tmp1 = make_pair(-1, -1), tmp2 = make_pair(-1, -1);
  if (mid >= x) tmp1 = query(o << 1, l, mid, x, y);
  if (mid < y) tmp2 = query(o << 1 | 1, mid + 1, r, x, y);
  if (tmp1.first == -1) return tmp2;
  if (tmp2.first == -1) return tmp1;
  if (tmp1.first > tmp2.first)
    return tmp2;
  else
    return tmp1;
}
void print() {
  for (int i = 1; i <= n; i++) {
    cout << ans[i] << " ";
  }
  cout << endl;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d%d", &L[i], &R[i]);
    v[L[i]].push_back(make_pair(R[i], i));
  }
  set<pair<int, int> > s;
  for (int i = 1; i <= n; i++) {
    s.insert(v[i].begin(), v[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, R[pos[i]]);
    if (i + 1 > R[pos[i]]) continue;
    if (tmp.first <= i) {
      printf("NO\n");
      print();
      swap(ans[pos[tmp.second]], ans[pos[i]]);
      print();
      return 0;
    }
  }
  printf("YES\n");
  print();
}