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 + 7;
const int K = 22;
namespace SP {
int p[N];
int sp[K][N];
int n;
vector<int> a;
void run() {
  for (int i = 0; i < n; i++) {
    sp[0][i] = i;
  }
  for (int i = 0; i + 1 < K; i++) {
    for (int j = 0; j + (2 << i) <= n; j++) {
      if (a[sp[i][j]] > a[sp[i][j + (1 << i)]]) {
        sp[i + 1][j] = sp[i][j + (1 << i)];
      } else {
        sp[i + 1][j] = sp[i][j];
      }
    }
  }
  p[1] = 0;
  for (int i = 2; i <= n; i++) {
    p[i] = p[i / 2] + 1;
  }
}
void init(vector<int> a_) {
  a = a_;
  n = (int)a.size();
  run();
}
int min(int l, int r) {
  int pw = p[r - l];
  int f = sp[pw][l];
  int s = sp[pw][r - (1 << pw)];
  if (a[f] < a[s])
    return f;
  else
    return s;
}
}  // namespace SP
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  vector<int> a(n);
  vector<int> b(n);
  vector<vector<pair<int, int>>> op(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i] >> b[i];
    a[i]--;
    b[i]--;
    op[a[i]].push_back({b[i], i});
  }
  vector<int> p(n);
  priority_queue<pair<int, int>, vector<pair<int, int>>,
                 greater<pair<int, int>>>
      q;
  for (int i = 0; i < n; i++) {
    for (auto t : op[i]) {
      q.push(t);
    }
    auto tp = q.top();
    q.pop();
    p[tp.second] = i;
    while (!q.empty() && q.top().first == i) {
      q.pop();
    }
  }
  vector<int> wr(n);
  vector<int> pa(n);
  for (int i = 0; i < n; i++) {
    pa[p[i]] = a[i];
    wr[p[i]] = i;
  }
  SP::init(pa);
  for (int i = 0; i < n; i++) {
    if (p[i] == b[i]) continue;
    int vl = SP::min(p[i] + 1, b[i] + 1);
    if (a[wr[vl]] <= p[i]) {
      cout << "NO\n";
      for (auto t : p) {
        cout << t + 1 << ' ';
      }
      cout << '\n';
      swap(p[i], p[wr[vl]]);
      for (auto t : p) {
        cout << t + 1 << ' ';
      }
      cout << '\n';
      return 0;
    }
  }
  cout << "YES\n";
  for (auto t : p) {
    cout << t + 1 << ' ';
  }
  cout << '\n';
}