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;
int n, m, ans, l[200005], r[200005], tree[800005];
pair<pair<int, int>, int> a[200005];
void c_tree(int l, int r, int p, int x, int y) {
  if (l == r)
    tree[p] = y;
  else {
    if (x <= ((l + r) / 2))
      c_tree(l, ((l + r) / 2), p * 2, x, y);
    else
      c_tree(((l + r) / 2) + 1, r, p * 2 + 1, x, y);
    tree[p] = max(tree[p * 2], tree[p * 2 + 1]);
  }
}
int q_tree(int l, int r, int p, int a, int b) {
  if (l == a && r == b)
    return tree[p];
  else {
    if (b <= ((l + r) / 2))
      return q_tree(l, ((l + r) / 2), p * 2, a, b);
    else if (a >= ((l + r) / 2) + 1)
      return q_tree(((l + r) / 2) + 1, r, p * 2 + 1, a, b);
    else
      return max(q_tree(l, ((l + r) / 2), p * 2, a, ((l + r) / 2)),
                 q_tree(((l + r) / 2) + 1, r, p * 2 + 1, ((l + r) / 2) + 1, b));
  }
}
bool cmp(pair<pair<int, int>, int> x, pair<pair<int, int>, int> y) {
  return l[x.second] < l[y.second];
}
set<pair<int, int>> s;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++)
    scanf("%d %d", &a[i].first.first, &a[i].first.second), a[i].second = i;
  sort(a + 1, a + 1 + n);
  for (int i = 1, i0 = 1; i <= n; i++) {
    while (i0 <= n && a[i0].first.first <= i)
      s.insert({a[i0].first.second, a[i0].second}), i0++;
    int p = s.begin()->second;
    s.erase(s.begin());
    l[p] = r[p] = i;
  }
  sort(a + 1, a + 1 + n, cmp);
  for (int i = 1; i <= n; i++) {
    if (q_tree(1, n, 1, a[i].first.first, i) >= i) {
      for (int i0 = a[i].first.first; i0 < i; i0++) {
        if (a[i0].first.second >= i) {
          swap(r[a[i].second], r[a[i0].second]);
          printf("NO\n");
          for (int i = 1; i <= n; i++)
            printf("%d%c", l[i], i == n ? '\n' : ' ');
          for (int i = 1; i <= n; i++)
            printf("%d%c", r[i], i == n ? '\n' : ' ');
          return 0;
        }
      }
    }
    c_tree(1, n, 1, i, a[i].first.second);
  }
  printf("YES\n");
  for (int i = 1; i <= n; i++) printf("%d%c", l[i], i == n ? '\n' : ' ');
  return 0;
}