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 = 2e5 + 50;
const int mod = 1e9 + 7;
long long qp(long long a, long long n) {
  long long res = 1;
  while (n > 0) {
    if (n & 1) res = res * a % mod;
    a = a * a % mod;
    n >>= 1;
  }
  return res;
}
template <class T>
inline bool scan(T& ret) {
  char c;
  int sgn;
  if (c = getchar(), c == EOF) return 0;
  while (c != '-' && (c < '0' || c > '9')) c = getchar();
  sgn = (c == '-') ? -1 : 1;
  ret = (c == '-') ? 0 : (c - '0');
  while (c = getchar(), c >= '0' && c <= '9') ret = ret * 10 + (c - '0');
  ret *= sgn;
  return 1;
}
int n;
set<pair<int, int> > st;
vector<pair<int, int> > a[maxn];
int res[maxn];
void pt(const vector<int>& x) {
  for (auto i : x) printf("%d ", i);
  printf("\n");
}
int main(int argc, char* argv[]) {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    int l, r;
    scanf("%d%d", &l, &r);
    a[l].emplace_back(r, i);
  }
  for (int i = 1; i <= n; ++i) {
    for (auto j : a[i]) st.emplace(j);
    res[st.begin()->second] = i;
    st.erase(st.begin());
  }
  for (int i = 1; i <= n; ++i) {
    for (auto j : a[i]) st.emplace(j);
    if (st.size() > 1 && res[next(st.begin())->second] <= st.begin()->first) {
      puts("NO");
      pt(vector<int>(res, res + n));
      swap(res[next(st.begin())->second], res[st.begin()->second]);
      pt(vector<int>(res, res + n));
      exit(0);
    }
    st.erase(st.begin());
  }
  puts("YES");
  pt(vector<int>(res, res + n));
  return 0;
}
