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;
int a[200005], b[200005];
int ret[200005];
vector<int> getRet(int n) {
  set<pair<int, int> > s;
  priority_queue<pair<int, pair<int, int> >,
                 vector<pair<int, pair<int, int> > >,
                 greater<pair<int, pair<int, int> > > >
      pq;
  set<int> rem;
  for (int i = 0; i < n; ++i) {
    pq.push(make_pair(b[i], make_pair(a[i], i)));
    s.insert(make_pair(a[i], i));
    rem.insert(i + 1);
  }
  while (!pq.empty()) {
    pair<int, int> tmp = pq.top().second;
    pq.pop();
    s.erase(tmp);
    int ind = *rem.lower_bound(tmp.first);
    ret[tmp.second] = ind;
    rem.erase(ind);
  }
  vector<int> out;
  for (int i = 0; i < n; ++i) {
    out.push_back(ret[i]);
  }
  return out;
}
int mapping[200005];
pair<int, int> seg[200005 * 4];
void init(int pos, int l, int r) {
  if (l == r) {
    seg[pos] = {a[mapping[l]], mapping[l]};
    return;
  }
  int mid = (l + r) / 2;
  init(pos * 2, l, mid);
  init(pos * 2 + 1, mid + 1, r);
  seg[pos] = min(seg[pos * 2], seg[pos * 2 + 1]);
}
pair<int, int> query(int pos, int l, int r, int ql, int qr) {
  if (r < ql || qr < l) {
    return {1e9, -1};
  }
  if (ql <= l && r <= qr) {
    return seg[pos];
  }
  int mid = (l + r) / 2;
  return min(query(pos * 2, l, mid, ql, qr),
             query(pos * 2 + 1, mid + 1, r, ql, qr));
}
void solve() {
  int n;
  scanf("%d ", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%d %d ", &a[i], &b[i]);
  }
  vector<int> rr = getRet(n);
  for (int i = 0; i < n; ++i) {
    mapping[rr[i]] = i;
  }
  init(1, 1, n);
  vector<int> vv;
  for (int i = 1; i <= n; ++i) {
    int mr = b[mapping[i]];
    if (i + 1 <= mr) {
      pair<int, int> q = query(1, 1, n, i + 1, mr);
      if (q.first <= i) {
        vv = rr;
        swap(vv[q.second], vv[mapping[i]]);
        break;
      }
    }
  }
  if (vv.size() == 0) {
    printf("YES\n");
    for (int x : rr) {
      printf("%d ", x);
    }
  } else {
    printf("NO\n");
    for (int x : rr) {
      printf("%d ", x);
    }
    putchar('\n');
    for (int x : vv) {
      printf("%d ", x);
    }
  }
}
int main() {
  solve();
  return 0;
}
