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 = 1e9 + 7;
const int MOD = 1e9 + 7;
const int dx[4] = {0, 0, -1, 1};
const int dy[4] = {-1, 1, 0, 0};
const int N = 2e5 + 1;
int pos[N];
struct Data {
  int x, y, pos;
  Data(int x, int y, int pos) : x(x), y(y), pos(pos) {}
};
int n;
vector<Data> F, G;
bool comp(Data A, Data B) { return (A.x == B.x) ? (A.y < B.y) : (A.x < B.x); }
vector<int> Solve(vector<Data> cur) {
  vector<int> res(n);
  priority_queue<pair<int, int>, vector<pair<int, int> >,
                 greater<pair<int, int> > >
      Pr;
  int p = 0;
  for (int i = 1; i <= n; i++) {
    while (p < n && cur[p].x <= i) {
      Pr.push(pair<int, int>(cur[p].y, cur[p].pos));
      p++;
    }
    int p = Pr.top().second;
    res[p] = i;
    Pr.pop();
    if (Pr.size()) {
      pos[p] = Pr.top().second;
    }
  }
  return res;
}
pair<int, int> memo[200001];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) pos[i] = -1;
  for (int i = 1; i <= n; i++) {
    int u, v, x, y;
    cin >> u >> v;
    F.push_back(Data(u, v, i - 1));
    memo[i - 1] = pair<int, int>(u, v);
  }
  sort(F.begin(), F.end(), comp);
  vector<int> x = Solve(F);
  int meow = -1;
  auto Check = [&](int val, pair<int, int> cur) {
    return cur.first <= val && val <= cur.second;
  };
  for (int i = 0; i < x.size(); i++) {
    if (pos[i] == -1) continue;
    int p = x[i];
    int q = x[pos[i]];
    if (i == pos[i]) continue;
    if (Check(q, memo[i]) && Check(p, memo[pos[i]])) {
      meow = i;
      break;
    }
  }
  if (meow == -1) {
    cout << "YES\n";
    for (int k : x) cout << k << ' ';
  } else {
    cout << "NO\n";
    for (int k : x) cout << k << ' ';
    cout << '\n';
    swap(x[meow], x[pos[meow]]);
    for (int k : x) cout << k << ' ';
  }
  return 0;
}