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 long long INF = 0x3f3f3f3f3f3f3f3fll;
const int mxN = 2e5 + 1;
pair<pair<long long, long long>, long long> a[mxN];
int ans1[mxN], ans2[mxN];
int rev_ans[mxN];
long long n;
long long seg[4 * mxN] = {};
void upd(long long i, long long val, long long st = 1, long long en = n,
         long long second = 0) {
  if (st <= i && i <= en) seg[second] = max(seg[second], val);
  if (i < st || i > en || st == en) return;
  long long mid = (st + en) / 2;
  upd(i, val, st, mid, 2 * second + 1);
  upd(i, val, mid + 1, en, 2 * second + 2);
}
long long qry(long long qs, long long qe, long long st = 1, long long en = n,
              long long second = 0) {
  if (qe < st || en < qs) return -1e18;
  if (qs <= st && en <= qe) return seg[second];
  long long mid = (st + en) / 2;
  return max(qry(qs, qe, st, mid, 2 * second + 1),
             qry(qs, qe, mid + 1, en, 2 * second + 2));
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i <= n - 1; i++)
    cin >> a[i].first.second >> a[i].first.first, a[i].second = i;
  sort(a, a + n);
  bool ok = 1;
  set<int> avail;
  for (int i = 1; i <= n; i++) avail.insert(i);
  for (int i = 0; i < n; i++) {
    auto it = avail.lower_bound(a[i].first.second);
    int cur = *it;
    avail.erase(it);
    ans1[a[i].second] = cur;
    ans2[a[i].second] = cur;
    rev_ans[cur] = i;
    if (ok) upd(cur, a[i].first.first);
    if (ok && cur != a[i].first.second &&
        qry(a[i].first.second, cur - 1) >= cur) {
      ok = 0;
      for (int j = a[i].first.second; j < cur; j++) {
        int i2 = rev_ans[j];
        if (a[i2].first.first >= cur) {
          ans2[a[i2].second] = cur;
          ans2[a[i].second] = j;
          break;
        }
      }
    }
  }
  if (ok) {
    cout << "YES\n";
    for (int i = 0; i <= n - 1; i++) cout << ans1[i] << " ";
    cout << '\n';
  } else {
    cout << "NO\n";
    for (int i = 0; i <= n - 1; i++) cout << ans1[i] << " ";
    cout << '\n';
    for (int i = 0; i <= n - 1; i++) cout << ans2[i] << " ";
    cout << '\n';
  }
}