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 double eps = 1e-9;
const int inf = 2000000000;
const long long infLL = 9000000000000000000;
template <typename first, typename second>
ostream& operator<<(ostream& os, const pair<first, second>& p) {
  return os << "(" << p.first << ", " << p.second << ")";
}
template <typename T>
ostream& operator<<(ostream& os, const vector<T>& v) {
  os << "{";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if (it != v.begin()) os << ", ";
    os << *it;
  }
  return os << "}";
}
template <typename T>
ostream& operator<<(ostream& os, const set<T>& v) {
  os << "[";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if (it != v.begin()) os << ",";
    os << *it;
  }
  return os << "]";
}
template <typename T>
ostream& operator<<(ostream& os, const multiset<T>& v) {
  os << "[";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if (it != v.begin()) os << ", ";
    os << *it;
  }
  return os << "]";
}
template <typename first, typename second>
ostream& operator<<(ostream& os, const map<first, second>& v) {
  os << "[";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if (it != v.begin()) os << ", ";
    os << it->first << " = " << it->second;
  }
  return os << "]";
}
void faltu() { cerr << '\n'; }
template <typename T>
void faltu(T a[], int n) {
  for (int i = 0; i < n; ++i) cerr << a[i] << ' ';
  cerr << '\n';
}
template <typename T, typename... hello>
void faltu(T arg, const hello&... rest) {
  cerr << arg << ' ';
  faltu(rest...);
}
const int mx = 2e5 + 5;
int n;
int a[mx], b[mx];
int arr[mx], rev[mx];
pair<int, int> tree[mx * 4];
int rev2[mx];
void init(int node, int l, int r) {
  if (l == r) {
    tree[node] = make_pair(a[arr[l]], l);
    return;
  }
  int mid = (l + r) / 2;
  init(2 * node, l, mid);
  init(2 * node + 1, mid + 1, r);
  tree[node] = min(tree[2 * node], tree[2 * node + 1]);
}
pair<int, int> query(int node, int l, int r, int b, int e) {
  if (l > e || r < b) return make_pair(inf, 0);
  if (l >= b && r <= e) return tree[node];
  int mid = (l + r) / 2;
  return min(query(2 * node, l, mid, b, e),
             query(2 * node + 1, mid + 1, r, b, e));
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> a[i] >> b[i];
  vector<pair<pair<int, int>, int> > vec;
  for (int i = 1; i <= n; ++i)
    vec.push_back(make_pair(make_pair(a[i], b[i]), i));
  sort(vec.begin(), vec.end());
  multiset<pair<pair<int, int>, int> > mst;
  for (int i = 0, j = 1; j <= n; ++j) {
    while (i < (int)vec.size() && vec[i].first.first <= j) {
      mst.insert(make_pair(make_pair(vec[i].first.second, vec[i].first.first),
                           vec[i].second));
      ++i;
    }
    auto x = *mst.begin();
    mst.erase(x);
    swap(x.first.first, x.first.second);
    int le = j;
    arr[le] = x.second;
    rev[x.second] = le;
  }
  for (int i = 1; i <= n; ++i) rev2[i] = rev[i];
  init(1, 1, n);
  bool ok = 0;
  for (int i = 1; i <= n; ++i) {
    pair<int, int> temp = query(1, 1, n, i + 1, b[arr[i]]);
    if (temp.first <= i) {
      ok = 1;
      swap(rev2[arr[i]], rev2[arr[temp.second]]);
      break;
    }
  }
  if (!ok) {
    cout << "YES" << '\n';
    for (int i = 1; i <= n; ++i) cout << rev[i] << " ";
    cout << '\n';
  } else {
    cout << "NO" << '\n';
    for (int i = 1; i <= n; ++i) cout << rev[i] << " ";
    cout << '\n';
    for (int i = 1; i <= n; ++i) cout << rev2[i] << " ";
    cout << '\n';
  }
}