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 = 210000;
vector<int> arr[maxn];
vector<int> parent(maxn);
vector<int> siz(maxn, 1);
vector<int> max_rep(maxn, -1);
int find_rep(int v) {
  if (parent[v] == v) return v;
  return parent[v] = find_rep(parent[v]);
}
void union_set(int a, int b) {
  int ra = find_rep(a);
  int rb = find_rep(b);
  if (ra != rb) {
    if (siz[ra] < siz[rb]) swap(ra, rb);
    parent[rb] = ra;
    int mx = max(max_rep[ra], max_rep[rb]);
    max_rep[ra] = max_rep[rb] = mx;
    siz[ra] += siz[rb];
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  for (int i = 1; i < maxn; i++) parent[i] = i, max_rep[i] = i;
  int n;
  cin >> n;
  vector<pair<int, int>> a(n);
  map<pair<int, int>, queue<int>> ans;
  for (auto& i : a)
    cin >> i.first >> i.second, arr[i.second].emplace_back(i.first);
  for (int i = 1; i <= n; i++) sort(arr[i].begin(), arr[i].end());
  vector<pair<int, int>> pr(n);
  vector<int> b(n);
  for (int i = 1; i <= n; i++)
    for (int& k : arr[i]) {
      int x = max_rep[find_rep(k)];
      union_set(x, x + 1);
      assert(x <= i);
      ans[{k, i}].push(x);
    }
  for (int i = 0; i < n; i++) {
    b[i] = ans[a[i]].front();
    ans[a[i]].pop();
    pr[b[i] - 1] = a[i];
  }
  map<pair<int, int>, int> mp;
  for (auto& k : a) {
    if (mp[k]) {
      cout << "NO\n";
      int index1 = -1, index2 = -1;
      for (int i = 0; i < n; i++) {
        if (a[i] == k) {
          if (index1 == -1) {
            index1 = i;
            continue;
          }
          if (index1 != -1 && index2 == -1)
            index2 = i;
          else
            break;
        }
      }
      for (int i = 0; i < n; i++) cout << b[i] << ' ';
      cout << '\n';
      swap(b[index1], b[index2]);
      for (int i = 0; i < n; i++) cout << b[i] << ' ';
      cerr << "\n\nTime elapsed: " << 1000.00 * clock() / CLOCKS_PER_SEC
           << " ms.\n";
      return 0;
    }
    mp[k]++;
  }
  bool flag = false;
  auto check = [](const pair<int, int>& p, int it) -> bool {
    return it >= p.first && it <= p.second;
  };
  int ni = -1;
  for (int i = 0; i < n - 1; i++) {
    if (check(pr[i + 1], i + 1) && check(pr[i], i + 2)) {
      flag = true;
      ni = i + 1;
      break;
    }
  }
  if (!flag)
    cout << "YES\n";
  else
    cout << "NO\n";
  for (int& i : b) cout << i << ' ';
  if (flag) {
    cout << '\n';
    for (int& i : b) {
      if (ni == i)
        cout << i + 1 << ' ';
      else if (ni + 1 == i)
        cout << i - 1 << ' ';
      else
        cout << i << ' ';
    }
  }
  cerr << "\n\nTime elapsed: " << 1000.00 * clock() / CLOCKS_PER_SEC
       << " ms.\n";
  return 0;
}