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;
const int maxn = 212345;
const int inf = 0x3f3f3f3f;
inline int lowbit(int x) { return x & -x; }
int n;
int fen[maxn], id[maxn], l[maxn], r[maxn], ans[maxn];
vector<pair<int, int> > G[maxn];
void upd(int x, int v) {
  while (x <= n) {
    fen[x] = min(fen[x], v);
    x += lowbit(x);
  }
}
int query(int x) {
  int ret = inf;
  while (x > 0) {
    ret = min(ret, fen[x]);
    x -= lowbit(x);
  }
  return ret;
}
int main() {
  scanf("%d", &n);
  for (int(i) = (1); (i) <= (n); (i)++) {
    scanf("%d%d", l + i, r + i);
    G[l[i]].emplace_back(r[i], i);
  }
  set<pair<int, int> > s;
  for (int(i) = (1); (i) <= (n); (i)++) {
    for (auto _ : G[i]) s.emplace(_);
    if (!s.empty() && s.begin()->first >= i) {
      ans[s.begin()->second] = i;
      id[i] = s.begin()->second;
      s.erase(s.begin());
    }
  }
  memset(fen, 0x3f, sizeof(fen));
  for (int(i) = (n); (i) >= (1); (i)--) {
    if (query(r[id[i]]) <= i)
      for (int(j) = (i + 1); (j) <= (n); (j)++)
        if (l[id[j]] <= i) {
          puts("NO");
          for (int(k) = (1); (k) <= (n); (k)++) printf("%d ", ans[k]);
          putchar('\n');
          swap(ans[id[i]], ans[id[j]]);
          for (int(k) = (1); (k) <= (n); (k)++) printf("%d ", ans[k]);
          return 0;
        }
    upd(i, l[id[i]]);
  }
  puts("YES");
  for (int(i) = (1); (i) <= (n); (i)++) printf("%d ", ans[i]);
}
