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;
const int N = 2e5 + 10;
int n;
int l[N], r[N];
int v[N], loc[N];
vector<pair<int, int> > w[N];
int f[N * 4];
void ins(int l, int r, int s, int ll, int v) {
  while (1) {
    f[s] = v;
    if (l == r) return;
    if ((l + r) / 2 >= ll)
      r = (l + r) / 2, s += s;
    else
      l = (l + r) / 2 + 1, s += s + 1;
  }
}
int get(int l, int r, int s, int ll, int rr) {
  if (!f[s] || r < ll || rr < l) return 0;
  if (ll <= l && r <= rr) return f[s];
  int x = get(l, (l + r) / 2, s + s, ll, rr);
  return x ? x : get((l + r) / 2 + 1, r, s + s + 1, ll, rr);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d %d", &l[i], &r[i]);
    w[l[i]].push_back({r[i], i});
  }
  priority_queue<pair<int, int> > h;
  for (int i = 1; i <= n; i++) {
    for (auto u : w[i]) h.push({-u.first, u.second});
    loc[v[i] = h.top().second] = i;
    h.pop();
  }
  for (int i = 1; i <= n; i++) w[i].clear();
  for (int i = 1; i <= n; i++) w[r[i]].push_back({loc[i], i});
  for (int i = n; i >= 1; i--) {
    for (auto u : w[i]) ins(1, n, 1, u.first, u.second);
    int x = get(1, n, 1, l[v[i]], i - 1);
    if (x) {
      puts("NO");
      for (int i = 1; i <= n; i++) printf("%d%c", loc[i], " \n"[i == n]);
      swap(loc[v[i]], loc[x]);
      for (int i = 1; i <= n; i++) printf("%d%c", loc[i], " \n"[i == n]);
      return 0;
    }
  }
  puts("YES");
  for (int i = 1; i <= n; i++) printf("%d%c", loc[i], " \n"[i == n]);
  return 0;
}
