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>
const int N = 2e5 + 5;
using namespace std;
int n, ans[N];
struct node {
  int id, l, r, c;
} p[N], b[N];
set<pair<int, int> > s;
set<pair<int, int> >::iterator it;
set<int> S;
set<int>::iterator it1;
template <typename T>
inline T read() {
  T x = 0, w = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') w = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
  return x * w;
}
bool cmp1(node a, node b) { return a.l < b.l; }
bool cmp2(node a, node b) { return a.c < b.c; }
int main() {
  n = read<int>();
  for (int i = 1; i <= n; i++)
    p[i].id = i, p[i].l = read<int>(), p[i].r = read<int>();
  sort(p + 1, p + n + 1, cmp1);
  for (int pos = 1, i = 1; i <= n; i++) {
    while (pos <= n && p[pos].l <= i)
      s.insert(make_pair(p[pos].r, p[pos].id)), pos++;
    it = s.begin(), ans[(*it).second] = i, s.erase(it);
  }
  while (s.size()) s.erase(s.begin());
  for (int i = 1; i <= n; i++) p[i].c = ans[p[i].id];
  for (int i = 1; i <= n; i++) b[i] = p[i];
  sort(b + 1, b + n + 1, cmp1);
  sort(p + 1, p + n + 1, cmp2);
  for (int pos = 1, i = 1; i <= n; i++) {
    while (pos <= n && b[pos].l <= i)
      s.insert(make_pair(b[pos].r, b[pos].c)), S.insert(b[pos].c), pos++;
    while (s.size()) {
      it = s.begin();
      if ((*it).first < i)
        S.erase((*it).second), s.erase(it);
      else
        break;
    }
    s.erase(make_pair(p[i].r, p[i].c)), S.erase(p[i].c);
    if (S.size()) {
      it1 = S.lower_bound(p[i].l);
      if (it1 != S.end() && (*it1) <= p[i].r) {
        puts("NO");
        for (int j = 1; j <= n; j++)
          printf("%d%c", ans[j], j == n ? '\n' : ' ');
        swap(ans[p[i].id], ans[p[(*it1)].id]);
        for (int j = 1; j <= n; j++)
          printf("%d%c", ans[j], j == n ? '\n' : ' ');
        return 0;
      }
    }
    s.insert(make_pair(p[i].r, p[i].c)), S.erase(p[i].c);
  }
  puts("YES");
  for (int j = 1; j <= n; j++) printf("%d%c", ans[j], j == n ? '\n' : ' ');
  return 0;
}
