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 MOD = 998244353, N = (int)2e5 + 50;
struct Tree {
  pair<int, int> dat[N * 4];
  void init_dat() { fill(dat, dat + N * 4, pair<int, int>{-1, -1}); }
  void update(int pos, int x, int l, int r, pair<int, int> p) {
    int mid = (l + r) / 2;
    if (l == r) {
      dat[x] = p;
      return;
    }
    if (pos <= mid)
      update(pos, 2 * x + 1, l, mid, p);
    else
      update(pos, 2 * x + 2, mid + 1, r, p);
    dat[x] = max(dat[2 * x + 1], dat[2 * x + 2]);
  }
  pair<int, int> query(int a, int b, int x, int l, int r) {
    if (r < a || b < l) return {-1, -1};
    int mid = (l + r) / 2;
    if (a <= l && r <= b) return dat[x];
    return max(query(a, b, 2 * x + 1, l, mid),
               query(a, b, 2 * x + 2, mid + 1, r));
  }
} tree;
struct rg {
  int l, r, idx, loc;
};
int n;
rg rgs[N];
int res[N];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  for (int i = 0; i < (n); ++i) {
    cin >> rgs[i].l >> rgs[i].r;
    rgs[i].l--, rgs[i].r--;
    rgs[i].idx = i;
  }
  set<int> S;
  for (int i = 0; i < (n); ++i) S.insert(i);
  sort(rgs, rgs + n, [](const rg &a, const rg &b) { return a.r < b.r; });
  for (int i = 0; i < (n); ++i) {
    auto it = S.lower_bound(rgs[i].l);
    assert(it != S.end());
    rgs[i].loc = *it;
    S.erase(it);
  }
  for (int i = 0; i < (n); ++i) res[rgs[i].idx] = rgs[i].loc;
  tree.init_dat();
  sort(rgs, rgs + n, [](const rg &a, const rg &b) { return a.loc < b.loc; });
  for (int i = 0; i < (n); ++i) {
    pair<int, int> p = tree.query(rgs[i].l, rgs[i].r, 0, 0, n - 1);
    if (p.first >= rgs[i].loc) {
      cout << "NO\n";
      for (int j = 0; j < (n); ++j) cout << res[j] + 1 << " ";
      cout << '\n';
      swap(res[rgs[i].idx], res[p.second]);
      for (int j = 0; j < (n); ++j) cout << res[j] + 1 << " ";
      cout << '\n';
      return 0;
    }
    tree.update(rgs[i].loc, 0, 0, n - 1, {rgs[i].r, rgs[i].idx});
  }
  cout << "YES\n";
  for (int j = 0; j < (n); ++j) cout << res[j] + 1 << " ";
  cout << '\n';
}