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 MAXN = 2e5 + 5;
struct Item {
  int idx, L, R;
  Item() = default;
  Item(int _i, int _L, int _R) : idx(_i), L(_L), R(_R) {}
  bool operator<(const Item& rhs) const {
    return (L == rhs.L) ? ((R == rhs.R) ? idx < rhs.idx : R < rhs.R)
                        : L < rhs.L;
  }
} C[MAXN];
int n;
int A[MAXN], B[MAXN];
set<int> Vertex;
multiset<pair<int, int> > S;
int pre[MAXN], vis[MAXN], Idx[MAXN], Pos[MAXN];
void MultiAns(int u, int nxt) {
  static int Cyc[MAXN], nC;
  puts("NO");
  for (int i = 1; i <= n; ++i) printf("%d%c", Idx[i], " \n"[i == n]);
  Cyc[nC = 1] = Pos[nxt];
  for (int v = u; v != Pos[nxt]; v = pre[v]) Cyc[++nC] = v;
  int v = Idx[Cyc[nC]];
  for (int i = 1; i <= nC; ++i) swap(v, Idx[Cyc[i]]);
  for (int i = 1; i <= n; ++i) printf("%d%c", Idx[i], " \n"[i == n]);
  exit(0);
}
void dfs(int u) {
  vector<int> Nxt;
  vis[u] = 1;
  while (true) {
    auto ite = Vertex.lower_bound(A[u]);
    if (ite != Vertex.end() && *ite == Idx[u]) Nxt.push_back(*ite), ++ite;
    if (ite == Vertex.end() || *ite > B[u]) break;
    int v = Pos[*ite];
    if (vis[v] == 1) return MultiAns(u, *ite);
    Nxt.push_back(*ite);
    if (vis[v] == 2) continue;
    pre[v] = u, dfs(v);
  }
  vis[u] = 2;
  for (const int& v : Nxt) Vertex.erase(v);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d%d", A + i, B + i);
  for (int i = 1; i <= n; ++i) C[i] = Item(i, A[i], B[i]);
  sort(C + 1, C + 1 + n);
  for (int p = 1, i = 1; i <= n; ++i) {
    while (p <= n && C[p].L <= i)
      S.insert(pair<int, int>(C[p].R, C[p].idx)), ++p;
    int d = S.begin()->second;
    Pos[i] = d, Idx[d] = i, S.erase(*S.begin());
  }
  for (int i = 1; i <= n; ++i) Vertex.insert(i);
  for (int i = 1; i <= n; ++i)
    if (!vis[i]) dfs(i);
  puts("YES");
  for (int i = 1; i <= n; ++i) printf("%d%c", Idx[i], " \n"[i == n]);
  return 0;
}