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 + 5;
int _w;
set<int> S;
priority_queue<pair<int, int> > Q;
int n, L[N], R[N], match[N], fr[N], vis[N];
void another(int u, int v) {
  static int rmp[N];
  puts("NO");
  for (int i = 1; i <= n; ++i) rmp[match[i]] = i;
  for (int i = 1; i <= n; ++i) printf("%d ", rmp[i]);
  putchar('\n');
  int x = u;
  while (x != v) x = match[rmp[x]] = fr[x];
  match[rmp[v]] = u;
  for (int i = 1; i <= n; ++i) rmp[match[i]] = i;
  for (int i = 1; i <= n; ++i) printf("%d ", rmp[i]);
  exit(0);
}
void dfs(int u) {
  set<int>::iterator it;
  vector<int> del;
  vis[u] = 1;
  int v;
  while (!S.empty()) {
    it = S.lower_bound(L[u]);
    if (it != S.end() && match[*it] == u) {
      del.push_back(*it);
      ++it;
    }
    if (it == S.end()) break;
    if (*it > R[u]) break;
    v = match[*it];
    if (vis[v] == 1) another(u, v);
    del.push_back(*it);
    if (vis[v] == 2) continue;
    fr[v] = u;
    dfs(v);
  }
  while (!del.empty()) {
    if (S.count(del.back())) S.erase(del.back());
    del.pop_back();
  }
  vis[u] = 2;
}
int main(void) {
  static int id[N];
  _w = scanf("%d", &n);
  for (int i = 1; i <= n; ++i)
    _w = scanf("%d%d", L + i, R + i), id[i] = i, S.insert(i);
  sort(id + 1, id + 1 + n, [](int a, int b) { return L[a] < L[b]; });
  for (int i = 1, cur = 1; i <= n; ++i) {
    while (cur <= n && L[id[cur]] == i)
      Q.push(pair<int, int>(-R[id[cur]], id[cur])), ++cur;
    match[i] = Q.top().second;
    Q.pop();
  }
  for (int i = 1; i <= n; ++i) id[match[i]] = i;
  for (int i = 1; i <= n; ++i)
    if (!vis[i]) dfs(i);
  puts("YES");
  for (int i = 1; i <= n; ++i) printf("%d ", id[i]);
  return 0;
}