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;
int it[800005];
int a[200005], b[200005];
int at[200005];
int pos[200005];
vector<pair<pair<int, int>, int> > ax;
void update(int id, int l, int r, int pos, int val) {
  if (l == r) {
    it[id] = val;
    return;
  }
  int mid = (l + r) / 2;
  if (pos > mid) {
    update(id * 2 + 1, mid + 1, r, pos, val);
  } else {
    update(id * 2, l, mid, pos, val);
  }
  it[id] = min(it[id * 2], it[id * 2 + 1]);
}
int get(int id, int l, int r, int u, int v) {
  if (l > v || r < u) {
    return 1000000000;
  }
  if (u <= l && v >= r) {
    return it[id];
  }
  int mid = (l + r) / 2;
  return min(get(id * 2, l, mid, u, v), get(id * 2 + 1, mid + 1, r, u, v));
}
void build_tree(int id, int l, int r) {
  if (l == r) {
    it[id] = l;
    return;
  }
  int mid = (l + r) / 2;
  build_tree(id * 2, l, mid);
  build_tree(id * 2 + 1, mid + 1, r);
  it[id] = min(it[id * 2], it[id * 2 + 1]);
}
signed main() {
  cin >> n;
  build_tree(1, 1, n);
  for (int i = 1; i <= n; i++) {
    cin >> a[i] >> b[i];
    ax.push_back({{b[i], a[i]}, i});
  }
  sort(ax.begin(), ax.end());
  for (auto i : ax) {
    int ok = get(1, 1, n, i.first.second, i.first.first);
    at[ok] = i.second;
    pos[i.second] = ok;
    update(1, 1, n, ok, 100000000);
  }
  memset(it, 0, sizeof(it));
  for (int i = 1; i <= n; i++) {
    update(1, 1, n, i, -b[at[i]]);
  }
  for (int i = 1; i <= n; i++) {
    if (a[at[i]] == i) continue;
    int ok = get(1, 1, n, a[at[i]], i - 1);
    if (abs(ok) >= i) {
      for (int j = a[at[i]]; j < i; j++) {
        if (b[at[j]] >= i) {
          cout << "NO\n";
          for (int k = 1; k <= n; k++) {
            cout << pos[k] << " ";
          }
          cout << endl;
          swap(pos[at[i]], pos[at[j]]);
          for (int k = 1; k <= n; k++) {
            cout << pos[k] << " ";
          }
          return 0;
        }
      }
    }
  }
  cout << "YES\n";
  for (int i = 1; i <= n; i++) {
    cout << pos[i] << " ";
  }
}