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;
void debug_out() { cerr << endl; }
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cerr << " " << to_string(H);
  debug_out(T...);
}
class DebugStream {
} LOG;
template <typename T>
DebugStream &operator<<(DebugStream &s, const T &) {
  return s;
}
mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
const int maxN = 2e5 + 9, maxV = 1e6 + 9, MOD = 1e9 + 7, SQ = 335, lg = 20,
          bs = 29;
int le[maxN], ri[maxN], n, rsp = -1, nxt[maxN];
vector<int> solve(int cnd) {
  vector<int> inds(n), ret(n);
  iota(inds.begin(), inds.end(), 0);
  sort(inds.begin(), inds.end(), [&](int a, int b) { return le[a] < le[b]; });
  auto compare = [&](int a, int b) {
    return ri[a] > ri[b] || (ri[a] == ri[b] && a > b);
  };
  priority_queue<int, vector<int>, decltype(compare)> pq(compare);
  int cr = 0;
  auto fn = [&]() {
    int tpp = pq.top();
    pq.pop();
    if (pq.size()) {
      nxt[tpp] = pq.top();
    }
    ret[tpp] = cr++;
  };
  for (int i = 0, j = 0; i < n; i = j) {
    while (le[inds[i]] > cr) fn();
    for (; j < n && le[inds[i]] == le[inds[j]]; j++) pq.push(inds[j]);
  }
  while (pq.size()) fn();
  return ret;
}
void prt(vector<int> vc) {
  for (auto el : vc) cout << el + 1 << ' ';
  cout << '\n';
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> le[i] >> ri[i];
    le[i]--;
    42;
    nxt[i] = -1;
  }
  auto uno = solve(1);
  for (int i = 0; i < n; i++) {
    if (nxt[i] != -1 && uno[nxt[i]] < ri[i]) {
      cout << "NO\n";
      prt(uno);
      swap(uno[i], uno[nxt[i]]);
      prt(uno);
      exit(0);
    }
  }
  cout << "YES\n";
  prt(uno);
  exit(0);
}
