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;
template <typename T1, typename T2>
inline void chkmin(T1 &x, T2 y) {
  if (x > y) x = y;
}
template <typename T1, typename T2>
inline void chkmax(T1 &x, T2 y) {
  if (x < y) x = y;
}
using ll = long long;
using ld = long double;
const string FILENAME = "input";
const int MAXN = 200228;
int n;
int a[MAXN], b[MAXN];
int c[MAXN];
vector<pair<int, int> > g[MAXN];
bool ok(int i, int val) { return a[i] <= val && val <= b[i]; }
int who[MAXN];
vector<int> ending[MAXN];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i] >> b[i];
    g[a[i]].push_back({b[i], i});
  }
  set<pair<int, int> > s;
  for (int i = 1; i <= n; i++) {
    for (auto x : g[i]) {
      s.insert(x);
    }
    auto x = *s.begin();
    s.erase(x);
    c[x.second] = i;
  }
  for (int i = 0; i < n; i++) {
    who[c[i]] = i;
    ending[b[i]].push_back(i);
  }
  for (int i = 1; i <= n; i++) {
    if (!s.empty()) {
      auto x = *s.rbegin();
      int t = x.second;
      if (x.first >= a[who[i]]) {
        assert(ok(t, i) && ok(who[i], c[t]));
      }
      if (ok(t, i) && ok(who[i], c[t])) {
        cout << "NO\n";
        for (int is = 0; is < n; is++) {
          cout << c[is] << ' ';
        }
        cout << '\n';
        swap(c[t], c[who[i]]);
        for (int is = 0; is < n; is++) {
          cout << c[is] << ' ';
        }
        cout << '\n';
        return 0;
      }
    }
    s.insert(make_pair(i, who[i]));
    for (auto x : ending[i]) {
      s.erase(make_pair(c[x], x));
    }
  }
  for (int i = 0; i < n - 1; i++) {
    if (ok(i, c[i + 1]) && ok(i + 1, c[i])) {
      cout << "NO\n";
      for (int is = 0; is < n; is++) {
        cout << c[is] << ' ';
      }
      cout << '\n';
      swap(c[i], c[i + 1]);
      for (int is = 0; is < n; is++) {
        cout << c[is] << ' ';
      }
      cout << '\n';
      return 0;
    }
  }
  cout << "YES\n";
  for (int i = 0; i < n; i++) {
    cout << c[i] << ' ';
  }
  cout << '\n';
}
