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 MAXN = 2e5 + 5;
int L[MAXN], R[MAXN];
int pos[MAXN], ans[MAXN];
vector<pair<int, int> > E[MAXN];
set<pair<int, int> > sv;
int n;
struct seg {
  int mi[MAXN << 2];
  void push_up(int rt) { mi[rt] = min(mi[rt << 1], mi[rt << 1 | 1]); }
  void build(int l, int r, int rt) {
    if (l == r) {
      mi[rt] = L[pos[l]];
      return;
    }
    int mid = (l + r) >> 1;
    build(l, mid, rt << 1);
    build(mid + 1, r, rt << 1 | 1);
    push_up(rt);
  }
  int query(int L, int R, int l, int r, int rt) {
    if (L > R) return 1e9;
    if (L <= l && r <= R) return mi[rt];
    int mid = (l + r) >> 1;
    int ret = 1e9;
    if (L <= mid) ret = min(ret, query(L, R, l, mid, rt << 1));
    if (mid < R) ret = min(ret, query(L, R, mid + 1, r, rt << 1 | 1));
    return ret;
  }
} se;
void print() {
  for (int i = 1; i <= n; i++) printf("%d ", ans[i]);
  puts("");
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d%d", &L[i], &R[i]);
    E[L[i]].push_back(make_pair(R[i], i));
  }
  for (int i = 1; i <= n; i++) {
    sv.insert(E[i].begin(), E[i].end());
    int id = (*sv.begin()).second;
    pos[i] = id;
    ans[id] = i;
    sv.erase(sv.begin());
  }
  se.build(1, n, 1);
  for (int i = 1; i <= n; i++) {
    int mi = se.query(i + 1, R[pos[i]], 1, n, 1);
    if (mi <= i) {
      puts("NO");
      for (int j = i + 1; j <= R[pos[i]]; j++)
        if (L[pos[j]] == mi) {
          print();
          swap(ans[pos[i]], ans[pos[j]]);
          print();
          return 0;
        }
    }
  }
  puts("YES");
  print();
  return 0;
}