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 maxn = 200009;
const long long MOD = 119 << 23 | 1;
class {
 public:
  vector<pair<int, int> > V[maxn];
  int a[maxn], b[maxn];
  int ans[maxn];
  int n;
  void solve() {
    cin >> n;
    for (int i = 1; i <= n; ++i) {
      cin >> a[i] >> b[i];
      V[a[i]].emplace_back(b[i], i);
    }
    priority_queue<pair<int, int> > que;
    for (int i = 1; i <= n; ++i) {
      for (auto j : V[i]) {
        que.emplace(-j.first, j.second);
      }
      ans[i] = que.top().second;
      que.pop();
    }
    build(1, 1, n);
    for (int i = 1; i <= n; ++i) {
      if (a[ans[i]] == i) continue;
      int res = query(1, 1, n, a[ans[i]], i - 1);
      if (res >= i) {
        cout << "NO\n";
        out();
        for (int j = a[ans[i]]; j <= i - 1; ++j) {
          if (b[ans[j]] >= i) {
            swap(ans[j], ans[i]);
            break;
          }
        }
        out();
        return;
      }
    }
    cout << "YES\n";
    out();
  }
  int c[maxn];
  void out() {
    for (int i = 1; i <= n; ++i) {
      c[ans[i]] = i;
    }
    for (int i = 1; i <= n; ++i) cout << c[i] << " ";
    cout << '\n';
  }
  int Max[maxn << 2];
  void build(int rt, int l, int r) {
    if (l == r) {
      Max[rt] = b[ans[l]];
      return;
    }
    int mid = l + r >> 1;
    build(rt << 1, l, mid);
    build(rt << 1 | 1, mid + 1, r);
    Max[rt] = max(Max[rt << 1], Max[rt << 1 | 1]);
  }
  int query(int rt, int l, int r, int ll, int rr) {
    if (ll <= l && rr >= r) return Max[rt];
    int mid = l + r >> 1;
    int res = 0;
    if (ll <= mid) res = max(res, query(rt << 1, l, mid, ll, rr));
    if (rr > mid) res = max(res, query(rt << 1 | 1, mid + 1, r, ll, rr));
    return res;
  }
} NSPACE;
int main() {
  ;
  ios_base::sync_with_stdio(false);
  cout.tie(0);
  cin.tie(0);
  NSPACE.solve();
}