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 int inf = 0x3f3f3f3f;
int N, a[200010], b[200010], ans[200010], pos[200010];
vector<pair<int, int> > d[200010];
int base;
pair<int, int> tree[1 << 19];
void init() {
  for (base = 1; base < N; base <<= 1)
    ;
  for (int i = 0; i < (1 << 19); i++) tree[i] = {inf, inf};
  for (int i = 1; i <= N; i++) tree[base + i - 1] = {a[pos[i]], pos[i]};
  for (int i = base - 1; i; i--) tree[i] = min(tree[i << 1], tree[i << 1 | 1]);
}
pair<int, int> query(int l, int r) {
  l += base - 1;
  r += base - 1;
  pair<int, int> ret = {inf, inf};
  while (l <= r) {
    if (l & 1) ret = min(ret, tree[l]);
    if (~r & 1) ret = min(ret, tree[r]);
    l = (l + 1) >> 1;
    r = (r - 1) >> 1;
  }
  return ret;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> N;
  for (int i = 1; i <= N; i++) {
    cin >> a[i] >> b[i];
    d[a[i]].push_back({b[i], i});
  }
  set<pair<int, int> > curr;
  for (int i = 1; i <= N; i++) {
    for (auto j : d[i]) curr.insert(j);
    pair<int, int> x = *curr.begin();
    curr.erase(curr.begin());
    ans[x.second] = i;
    pos[i] = x.second;
  }
  init();
  for (int i = 1; i <= N; i++) {
    pair<int, int> x = query(i + 1, b[pos[i]]);
    if (x.first <= i) {
      cout << "NO" << '\n';
      for (int i = 1; i <= N; i++) cout << ans[i] << ' ';
      cout << '\n';
      swap(ans[pos[i]], ans[x.second]);
      for (int i = 1; i <= N; i++) cout << ans[i] << ' ';
      cout << '\n';
      return 0;
    }
  }
  cout << "YES" << '\n';
  for (int i = 1; i <= N; i++) cout << ans[i] << ' ';
  cout << '\n';
  return 0;
}