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 
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
template <class T>
struct sparse {
  int n, h;
  vector<vector<T>> table;
  function<T(T, T)> func;
  sparse(vector<T> a, function<T(T, T)> f) {
    n = a.size();
    func = f;
    h = 32 - __builtin_clz(n);
    table.resize(h);
    table[0] = a;
    for (int j = 1; j < h; j++) {
      table[j].resize(n - (1 << j) + 1);
      for (int i = 0; i <= n - (1 << j); i++) {
        table[j][i] = func(table[j - 1][i], table[j - 1][i + (1 << (j - 1))]);
      }
    }
  }
  T get(int from, int to) const {
    int k = 32 - __builtin_clz(to - from + 1) - 1;
    return func(table[k][from], table[k][to - (1 << k) + 1]);
  }
};
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  vector<int> a(n), b(n);
  vector<vector<int>> p(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i] >> b[i];
    a[i]--, b[i]--;
    p[a[i]].emplace_back(i);
  }
  priority_queue<pair<int, int>, vector<pair<int, int>>,
                 greater<pair<int, int>>>
      pq;
  vector<int> ans(n);
  vector<int> c(n), l(n), r(n);
  for (int i = 0; i < n; i++) {
    for (int j : p[i]) {
      pq.emplace(b[j], j);
    }
    pair<int, int> t = pq.top();
    pq.pop();
    ans[t.second] = i;
    c[i] = t.second;
    l[i] = a[t.second];
    r[i] = b[t.second];
  }
  sparse<int> s(l, [&](int a, int b) { return min(a, b); });
  sparse<int> t(r, [&](int a, int b) { return max(a, b); });
  ;
  for (int i = 0; i < n; i++) {
    pair<int, int> sw = {-1, i};
    if (i + 1 <= r[i] && s.get(i + 1, r[i]) <= i) {
      for (int j = i + 1; j <= r[i]; j++) {
        if (l[j] <= i) {
          sw.first = j;
          break;
        }
      }
    }
    if (i <= i - 1 && t.get(l[i], i - 1) >= i) {
      for (int j = i - 1; j >= l[i]; j--) {
        if (r[j] >= i) {
          sw.first = j;
          break;
        }
      }
    }
    if (sw.first != -1) {
      ;
      cout << "NO" << '\n';
      for (int j : ans) {
        cout << j + 1 << " ";
      }
      cout << '\n';
      swap(ans[c[sw.first]], ans[c[sw.second]]);
      for (int j : ans) {
        cout << j + 1 << " ";
      }
      cout << '\n';
      return 0;
    }
  }
  cout << "YES" << '\n';
  for (int i : ans) {
    cout << i + 1 << " ";
  }
  cout << '\n';
  return 0;
}
