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 = 2e5 + 7, lim = 4 * maxn;
pair<int, int> segtree[lim];
set<pair<int, int> > s;
vector<pair<int, pair<int, int> > > v;
int ip[maxn][2], ans[maxn], pos[maxn], n, swapl = -1, swapr = -1;
void build(int cur, int l, int r) {
  if (l == r) {
    segtree[cur] = {ip[pos[l]][0], l};
    return;
  }
  int lc = 2 * cur, rc = lc + 1, mid = (l + r) / 2;
  build(lc, l, mid);
  build(rc, mid + 1, r);
  segtree[cur] = min(segtree[lc], segtree[rc]);
}
pair<int, int> query(int cur, int l, int r, int ql, int qr) {
  if (l > qr || r < ql || ql > qr) {
    return {INT_MAX, INT_MAX};
  }
  if (l >= ql && r <= qr) {
    return segtree[cur];
  }
  int lc = 2 * cur, rc = lc + 1, mid = (l + r) / 2;
  return min(query(lc, l, mid, ql, qr), query(rc, mid + 1, r, ql, qr));
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> ip[i][0] >> ip[i][1];
    v.push_back({ip[i][0], {ip[i][1], i}});
  }
  sort(v.rbegin(), v.rend());
  for (int i = 1; i <= n; i++) {
    while (!v.empty() && v.back().first <= i) {
      s.insert(v.back().second);
      v.pop_back();
    }
    pair<int, int> cur = *s.begin();
    s.erase(s.begin());
    ans[cur.second] = i;
    pos[i] = cur.second;
  }
  build(1, 1, n);
  for (int i = 1; i < n; i++) {
    int j = query(1, 1, n, i + 1, ip[pos[i]][1]).second;
    if (j == INT_MAX || ip[pos[j]][0] > i) continue;
    swapl = i;
    swapr = j;
  }
  if (swapl == -1) {
    cout << "YES\n";
  } else {
    cout << "NO\n";
  }
  for (int i = 1; i <= n; i++) cout << ans[i] << " ";
  cout << '\n';
  if (swapl != -1) {
    swap(ans[pos[swapl]], ans[pos[swapr]]);
    for (int i = 1; i <= n; i++) cout << ans[i] << " ";
  }
  return 0;
}