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;
int N;
pair<int, int> A[200005];
int Res[200005], Place[200005], Tree[800005];
set<pair<int, int> > X;
vector<int> V[200005];
void Read() {
  scanf("%d", &N);
  for (int i = 1; i <= N; i++) {
    scanf("%d%d", &A[i].first, &A[i].second);
    V[A[i].first].push_back(i);
  }
}
void findPerfectMatching() {
  for (int i = 1; i <= N; i++) {
    for (int j = 0; j < V[i].size(); j++) {
      int friendIndex = V[i][j];
      X.insert(make_pair(A[friendIndex].second, friendIndex));
    }
    int index = X.begin()->second;
    Res[index] = i;
    Place[i] = index;
    X.erase(X.begin());
  }
}
void buildTree(int K, int L, int R) {
  if (L == R) {
    Tree[K] = A[Place[L]].first;
    return;
  }
  buildTree(K * 2, L, (L + R) / 2);
  buildTree(K * 2 + 1, (L + R) / 2 + 1, R);
  Tree[K] = min(Tree[K * 2], Tree[K * 2 + 1]);
}
int Query(int K, int L, int R, int x, int y) {
  if (L > R || L > y || R < x) return N + 5;
  if (L >= x && R <= y) {
    return Tree[K];
  }
  return min(Query(K * 2, L, (L + R) / 2, x, y),
             Query(K * 2 + 1, (L + R) / 2 + 1, R, x, y));
}
void print2Solutions(int left) {
  int right;
  for (int i = left + 1; i <= A[Place[left]].second; i++) {
    if (A[Place[i]].first <= left) {
      right = i;
      break;
    }
  }
  printf("NO\n");
  for (int i = 1; i <= N; i++) printf("%d ", Res[i]);
  printf("\n");
  swap(Res[Place[left]], Res[Place[right]]);
  for (int i = 1; i <= N; i++) printf("%d ", Res[i]);
  printf("\n");
}
void print1Solution() {
  printf("YES\n");
  for (int i = 1; i <= N; i++) printf("%d ", Res[i]);
  printf("\n");
}
void Solve() {
  for (int i = 1; i <= N; i++) {
    int last = A[Place[i]].second;
    int minLeftHead = Query(1, 1, N, i + 1, last);
    if (minLeftHead > i) continue;
    print2Solutions(i);
    return;
  }
  print1Solution();
}
int main() {
  Read();
  findPerfectMatching();
  buildTree(1, 1, N);
  Solve();
  return 0;
}