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 ll = long long;
const int MN = 2e5 + 100;
int N;
struct endpt {
 public:
  int i, r;
  bool operator<(endpt o) const { return r < o.r || !(o.r < r) && i < o.i; }
};
std::vector<endpt> a[MN];
int ib[MN], b[MN], s1 = -1, s2;
const int MS = 1 << 19;
ll v[MS];
void upd(int n, int l, int r, int x, ll k) {
  if (r - l > 1) {
    int m = l + (r - l) / 2;
    if (x < m)
      upd(n << 1, l, m, x, k);
    else
      upd(n << 1 | 1, m, r, x, k);
    v[n] = std::max(v[n << 1], v[n << 1 | 1]);
  } else
    v[n] = k;
}
ll qry(int n, int l, int r, int ql, int qr) {
  if (ql <= l && r <= qr)
    return v[n];
  else {
    int m = l + (r - l) / 2;
    ll f = -1;
    if (ql < m) f = std::max(f, qry(n << 1, l, m, ql, qr));
    if (m < qr) f = std::max(f, qry(n << 1 | 1, m, r, ql, qr));
    return f;
  }
}
std::set<endpt> active;
int main(void) {
  scanf("%d", &N);
  for (int i = 0, x, y; i < N; ++i) {
    scanf("%d%d", &x, &y);
    a[x].push_back({i, ++y});
  }
  for (int i = 1; i <= N; ++i) {
    for (auto x : a[i]) active.insert(x);
    auto it = active.begin();
    auto y = *it;
    active.erase(it);
    b[y.i] = i;
    ib[i] = y.i;
  }
  memset(v, 0xff, sizeof(v));
  for (int i = 1; i <= N; ++i)
    for (auto x : a[i]) {
      ll z = qry(1, 1, N + 1, i, x.r), k = b[x.i];
      if (z >> 32 > k) {
        s1 = x.i, s2 = (int)(z & 0xffffffff);
        goto done;
      }
      upd(1, 1, N + 1, k, (ll)x.r << 32 | x.i);
    }
done:
  if (!~s1) {
    printf("YES\n");
    for (int i = 0; i < N; ++i) printf("%d%c", b[i], " \n"[i + 1 == N]);
  } else {
    printf("NO\n");
    for (int i = 0; i < N; ++i) printf("%d%c", b[i], " \n"[i + 1 == N]);
    std::swap(b[s1], b[s2]);
    for (int i = 0; i < N; ++i) printf("%d%c", b[i], " \n"[i + 1 == N]);
  }
  return 0;
}