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;
#pragma GCC optimize("O2")
#pragma GCC optimize("unroll-loops")
const long long maxn = 2e5 + 10, lg = 17, mod = 1e9 + 7, inf = 1e18;
long long n, a[maxn], b[maxn], we[maxn], ans[maxn];
vector<pair<long long, long long> > s[maxn];
pair<long long, long long> seg[maxn << 1];
void goz(long long i, pair<long long, long long> x) {
  seg[i += n] = x;
  for (; i > 1; i >>= 1) seg[i >> 1] = min(seg[i], seg[i ^ 1]);
}
long long get(long long l, long long r) {
  pair<long long, long long> ret = {inf, inf};
  for (l += n, r += n; l < r; l >>= 1, r >>= 1) {
    if (l & 1) ret = min(ret, seg[l++]);
    if (r & 1) ret = min(ret, seg[--r]);
  }
  return ret.second;
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  for (int i = 0; i < (maxn << 1); i++) seg[i] = {inf, inf};
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i] >> b[i];
    s[a[i]].push_back({b[i], i});
  }
  set<pair<long long, long long> > nw;
  for (int i = 1; i <= n; i++) {
    for (auto x : s[i]) nw.insert(x);
    long long xxx = (*nw.begin()).second;
    ans[xxx] = i, we[i] = xxx;
    nw.erase(nw.begin());
  }
  for (int i = 0; i < n; i++) goz(i, {a[we[i + 1]], i + 1});
  for (int i = 1; i <= n; i++) {
    long long j = get(i, b[we[i]]);
    if (j == inf) continue;
    if (a[we[j]] <= i) {
      cout << "NO\n";
      for (int k = 1; k <= n; k++) cout << ans[k] << ' ';
      cout << endl;
      swap(ans[we[i]], ans[we[j]]);
      for (int k = 1; k <= n; k++) cout << ans[k] << ' ';
      cout << endl;
      return 0;
    }
  }
  cout << "YES\n";
  for (int k = 1; k <= n; k++) cout << ans[k] << ' ';
  cout << endl;
  return 0;
}
