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 
4
3 4
1 2
3 4
1 2
 4
2 4
1 3
4 4
1 1
 10
6 6
1 2
2 2
6 7
8 8
3 5
5 5
4 4
8 9
10 10
 4
1 4
1 4
2 4
1 2
 8
3 4
7 8
2 6
1 3
6 7
5 8
4 6
8 8
 7
7 7
2 3
2 3
5 7
5 6
4 4
1 2
 15
2 2
10 11
2 5
6 7
7 12
9 9
2 3
5 14
1 3
10 10
15 15
12 12
5 6
6 6
14 15
 5
1 3
4 4
5 5
1 4
1 5
 4
1 4
2 3
2 3
4 4
 4
1 4
1 4
1 2
1 4
 5
1 3
1 3
1 5
3 4
1 5
 3
1 3
2 2
1 3
 5
4 4
1 4
3 4
1 3
3 5
 3
1 3
1 2
2 3
 4
2 4
1 3
2 3
1 2
 5
4 5
4 5
1 3
1 2
2 4
 6
2 6
4 6
6 6
2 2
3 3
1 3
 3
2 3
1 2
1 3
 6
4 6
1 3
4 6
4 4
1 4
1 2
 7
3 6
6 6
5 7
2 4
2 3
1 2
3 5
 8
4 6
8 8
1 4
4 5
6 8
1 2
3 6
1 4
 6
2 4
2 6
4 5
2 5
6 6
1 4
 8
4 4
7 7
3 5
5 7
1 8
4 5
1 5
2 2
 4
3 4
1 3
2 3
1 2
 10
9 9
3 4
1 1
5 5
7 7
3 7
6 10
3 3
10 10
1 3
 4
1 4
4 4
2 3
1 1
 1
1 1
 8
1 5
6 8
1 3
4 8
1 1
3 3
6 6
8 8
 5
3 3
4 4
2 2
1 1
5 5
 4
1 1
2 3
2 3
1 4
 4
3 4
1 4
2 3
1 2
 2
1 2
1 2
 8
3 3
3 5
2 3
1 1
5 7
7 7
5 5
3 8
 5
2 2
3 3
1 2
1 5
4 4
 6
1 3
2 4
3 4
2 3
5 5
6 6
 6
1 4
3 6
6 6
3 5
1 5
2 5
 5
4 4
2 3
4 5
1 4
2 4
 3
1 2
1 2
1 3
 4
1 4
2 2
3 3
1 4
 5
2 5
1 1
2 3
3 4
5 5
 4
2 4
1 4
3 4
1 3
 6
1 1
4 6
3 6
1 3
3 4
3 6
 5
2 3
3 4
4 5
1 1
2 3
 5
2 3
1 5
2 4
3 5
4 4
 20
12 14
13 13
18 18
3 3
7 8
11 11
10 13
2 2
9 10
3 5
7 8
15 17
14 15
9 9
20 20
15 16
6 6
19 19
1 1
5 5
 5
1 5
2 2
3 3
4 4
1 5
 7
5 7
1 5
1 2
1 5
5 6
3 5
7 7

NO
3 1 4 2
3 2 4 1
 NO
3 2 4 1 
2 3 4 1  YES
6 1 2 7 8 3 5 4 9 10  NO
2 3 4 1 
1 3 4 2 
 YES
3 7 2 1 6 5 4 8  NO
7 2 3 6 5 4 1 
7 3 2 6 5 4 1 
 YES
2 11 4 7 8 9 3 13 1 10 15 12 5 6 14  NO
1 4 5 2 3 
2 4 5 1 3  NO
1 2 3 4 
1 3 2 4 
 NO
2 3 1 4 
1 3 2 4 
 NO
1 2 4 3 5 
2 1 4 3 5 
 NO
1 2 3 
3 2 1 
 NO
4 2 3 1 5 
4 1 3 2 5  NO
2 1 3
1 2 3
 NO
4 2 3 1
4 1 3 2
 NO
4 5 2 1 3
4 5 1 2 3
 NO
4 5 6 2 3 1 
5 4 6 2 3 1 
 NO
2 1 3 
3 1 2 
 NO
5 2 6 4 3 1 
5 1 6 4 3 2 
 NO
5 6 7 3 2 1 4 
5 6 7 2 3 1 4  NO
5 8 2 4 7 1 6 3
5 8 1 4 7 2 6 3
 NO
2 5 4 3 6 1 
3 5 4 2 6 1  YES
4 7 3 6 8 5 1 2  NO
4 2 3 1
4 1 3 2
 YES
9 4 1 5 7 6 8 3 10 2  NO
3 4 2 1 
2 4 3 1  YES
1  NO
4 7 2 5 1 3 6 8 
5 7 2 4 1 3 6 8  YES
3 4 2 1 5  NO
1 2 3 4 
1 3 2 4 
 NO
3 4 2 1 
4 3 2 1 
 NO
1 2 
2 1 
 YES
3 4 2 1 6 7 5 8  YES
2 3 1 5 4  NO
1 3 4 2 5 6
1 2 4 3 5 6
 NO
1 5 6 3 2 4 
2 5 6 3 1 4 
 NO
4 2 5 1 3 
4 3 5 1 2  NO
1 2 3 
2 1 3 
 NO
1 2 3 4 
4 2 3 1 
 NO
4 1 2 3 5 
3 1 2 4 5 
 NO
2 3 4 1
2 1 4 3
 NO
1 4 5 2 3 6 
1 5 4 2 3 6 
 NO
2 4 5 1 3 
3 4 5 1 2 
 NO
2 1 3 5 4 
3 1 2 5 4  NO
14 13 18 3 7 11 12 2 10 4 8 17 15 9 20 16 6 19 1 5 
14 13 18 3 8 11 12 2 10 4 7 17 15 9 20 16 6 19 1 5 
 NO
1 2 3 4 5 
5 2 3 4 1 
 NO
6 2 1 3 5 4 7
6 1 2 3 5 4 7

#include <bits/stdc++.h>
using namespace std;
mt19937 __MT(chrono::system_clock::now().time_since_epoch().count());
const int maxn = 200000 + 10;
pair<int, int> itvl[maxn];
int n;
int id[maxn];
int ans[maxn];
int inv[maxn];
struct segtree {
  int maxv[maxn << 2];
  int maxid[maxn << 2];
  int n, ql, qr, idx, rm;
  void init(int n) {
    this->n = n;
    build(1, n, 1);
  }
  void maintain(int l, int r, int o) {
    if (l != r) {
      maxv[o] = max(maxv[o << 1], maxv[o << 1 | 1]);
      if (maxv[o << 1] == maxv[o]) maxid[o] = maxid[o << 1];
      if (maxv[o << 1 | 1] == maxv[o]) maxid[o] = maxid[o << 1 | 1];
    }
  }
  void build(int l, int r, int o) {
    if (l == r) {
      maxv[o] = itvl[ans[l]].second;
      maxid[o] = l;
      return;
    }
    int mid = l + r >> 1;
    build(l, mid, o << 1), build(mid + 1, r, o << 1 | 1);
    maintain(l, r, o);
  }
  void update(int o) {
    if (rm < maxv[o]) {
      rm = maxv[o];
      idx = maxid[o];
    }
  }
  void query(int l, int r, int o) {
    if (ql <= l && qr >= r)
      update(o);
    else {
      int mid = l + r >> 1;
      if (ql <= mid) query(l, mid, o << 1);
      if (qr > mid) query(mid + 1, r, o << 1 | 1);
    }
  }
  pair<int, int> Query(int l, int r) {
    rm = -1;
    idx = -1;
    ql = l, qr = r;
    query(1, n, 1);
    return (pair<int, int>){rm, idx};
  }
};
segtree tree;
inline void solve() {
  set<int> o;
  for (auto i = (1); i <= (n); ++i) o.insert(i);
  for (auto i = (1); i <= (n); ++i) {
    int j = id[i];
    auto it = o.lower_bound(itvl[j].first);
    ans[*it] = j;
    int pos = *it;
    o.erase(pos);
  }
  tree.init(n);
  pair<int, int> swp;
  int amb = 0;
  swp.first = swp.second = -1;
  for (auto i = (2); i <= (n); ++i) {
    int j = ans[i];
    assert(itvl[j].first <= i);
    auto e = tree.Query(itvl[j].first, i - 1);
    if (e.first >= i) {
      swp.first = i;
      swp.second = e.second;
      amb = 1;
      break;
    }
  }
  if (!amb) {
    puts("YES");
  } else {
    puts("NO");
  }
  for (auto i = (1); i <= (n); ++i) inv[ans[i]] = i;
  for (auto i = (1); i <= (n); ++i) printf("%d ", inv[i]);
  puts("");
  if (amb) {
    swap(ans[swp.first], ans[swp.second]);
    for (auto i = (1); i <= (n); ++i) inv[ans[i]] = i;
    for (auto i = (1); i <= (n); ++i) printf("%d ", inv[i]);
    puts("");
  }
}
int main(void) {
  scanf("%d", &(n));
  for (auto i = (1); i <= (n); ++i)
    scanf("%d%d", &itvl[i].first, &itvl[i].second), id[i] = i;
  sort(id + 1, id + n + 1, [](const int& a, const int& b) {
    int len[2];
    len[0] = itvl[a].second - itvl[a].first;
    len[1] = itvl[b].second - itvl[b].first;
    if (len[0] != len[1]) return len[0] < len[1];
    if (itvl[a].first != itvl[b].first) return itvl[a].first < itvl[b].first;
    return a < b;
  });
  solve();
  return 0;
}
