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;
const int Max_N = 1e5;
int n, p[Max_N + 10] = {}, pos[Max_N + 10] = {};
struct kk {
  int l, r, Id;
} a[Max_N + 10] = {};
struct Tree {
  int x, l, r, Min, Val;
} A[Max_N * 4 + 10] = {};
inline void Build(int x, int l, int r) {
  A[x].l = l;
  A[x].r = r;
  if (l == r) {
    A[x].Min = l;
    return;
  }
  Build(x << 1, l, (l + r) / 2);
  Build((x << 1) + 1, (l + r) / 2 + 1, r);
  A[x].Min = min(A[x << 1].Min, A[(x << 1) + 1].Min);
}
inline void Add(int x, int k) {
  if (A[x].l == A[x].r) {
    A[x].Min = 1 << 30;
    return;
  }
  if (k <= A[x << 1].r)
    Add(x << 1, k);
  else
    Add((x << 1) + 1, k);
  A[x].Min = (A[x << 1].Min, A[(x << 1) + 1].Min);
}
inline int Ask(int x, int l, int r) {
  if (l <= A[x].l && A[x].r <= r) return A[x].Min;
  if (r <= A[x << 1].r)
    return Ask(x << 1, l, r);
  else if (l > A[x << 1].r)
    return Ask((x << 1) + 1, l, r);
  else
    return min(Ask(x << 1, l, r), Ask((x << 1) + 1, l, r));
}
inline bool mycmp(kk x, kk y) { return x.r == y.r ? x.l < y.l : x.r < y.r; }
inline void BUILD(int x, int l, int r) {
  A[x].l = l;
  A[x].r = r;
  if (l == r) {
    A[x].Min = a[pos[l]].l;
    A[x].Val = a[pos[l]].Id;
    return;
  }
  BUILD(x << 1, l, (l + r) / 2);
  BUILD((x << 1) + 1, (l + r) / 2 + 1, r);
  A[x].Min = min(A[x << 1].Min, A[(x << 1) + 1].Min);
  A[x].Val =
      A[x << 1].Min < A[(x << 1) + 1].Min ? A[x << 1].Val : A[(x << 1) + 1].Val;
}
inline pair<int, int> ASK(int x, int l, int r) {
  if (l <= A[x].l && A[x].r <= r) return make_pair(A[x].Min, A[x].Val);
  pair<int, int> ans1 = make_pair(n + 1, 0), ans2 = make_pair(n + 1, 0);
  if (l <= A[x << 1].r) ans1 = ASK(x << 1, l, r);
  if (r >= A[(x << 1) + 1].r) ans2 = ASK((x << 1) + 1, l, r);
  return make_pair(min(ans1.first, ans2.first),
                   ans1.first < ans2.first ? ans1.second : ans2.second);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d%d", &a[i].l, &a[i].r), a[i].Id = i;
  Build(1, 1, n);
  sort(a + 1, a + n + 1, mycmp);
  for (int i = 1; i <= n; i++) {
    int x = Ask(1, a[i].l, a[i].r);
    pos[p[a[i].Id] = x] = i;
    Add(1, x);
  }
  int X = 0, Y = 0;
  BUILD(1, 1, n);
  for (int i = 1; i <= n; i++) {
    if (p[a[i].Id] == a[i].r) continue;
    pair<int, int> b = ASK(1, p[a[i].Id] + 1, a[i].r);
    if (b.first <= p[a[i].Id] && p[b.second] > p[a[i].Id]) {
      X = a[i].Id;
      Y = b.second;
      break;
    }
  }
  printf(X ? "NO\n" : "YES\n");
  for (int i = 1; i <= n; i++) printf("%d ", p[i]);
  puts("");
  if (X) {
    swap(p[X], p[Y]);
    for (int i = 1; i <= n; i++) printf("%d ", p[i]);
    puts("");
  }
  return 0;
}
