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;
struct data {
  int X, Y, Z;
};
const int N = (int)1e6 + 7;
const int MOD = (int)1e9 + 7;
const long long INF = (long long)1e18 + 7;
int IT[N];
data a[N], b[N];
int n;
bool cmp(data a, data b) { return (a.X != b.X) ? a.X < b.X : a.Y < b.Y; }
void Insert(int i, int l, int r, int x, int val) {
  if (x < l || x > r) return;
  if (l == r) {
    IT[i] = val;
    return;
  }
  int m = (l + r) >> 1;
  Insert(i * 2, l, m, x, val);
  Insert(i * 2 + 1, m + 1, r, x, val);
  IT[i] = max(IT[i * 2], IT[i * 2 + 1]);
  return;
}
int Get(int i, int l, int r, int L, int R) {
  if (l >= L && r <= R) return IT[i];
  if (L > r || R < l) return 0;
  int m = (l + r) >> 1;
  int x = Get(i * 2, l, m, L, R);
  int y = Get(i * 2 + 1, m + 1, r, L, R);
  return max(x, y);
}
void show() {
  vector<int> ans(n + 1);
  for (int i = 1; i <= n; i++) ans[a[i].Z] = i;
  for (int i = 1; i <= n; i++) cout << ans[i] << " ";
}
void builda() {
  priority_queue<pair<int, int>, vector<pair<int, int>>,
                 greater<pair<int, int>>>
      q;
  sort(a + 1, a + n + 1, cmp);
  int l = 1;
  for (int i = 1; i <= n; i++) {
    while (a[l].X == i && l <= n) q.push({a[l].Y, l++});
    pair<int, int> gt = q.top();
    q.pop();
    b[i] = a[gt.second];
  }
  for (int i = 1; i <= n; i++) a[i] = b[i];
  return;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i].X >> a[i].Y;
    a[i].Z = i;
  }
  builda();
  Insert(1, 1, n, 1, a[1].Y);
  int l = 0;
  for (int i = 2; i <= n; i++) {
    int R = i, L = a[i].X;
    int ck = Get(1, 1, n, L, R);
    Insert(1, 1, n, i, a[i].Y);
    if (ck >= i) {
      l = i;
      break;
    }
  }
  int r = 0;
  for (int i = a[l].X; i < l; i++)
    if (a[i].Y >= l) {
      r = i;
      break;
    }
  if (l) {
    cout << "NO\n";
    show();
    swap(a[l], a[r]);
    cout << "\n";
    show();
    return 0;
  }
  cout << "YES\n";
  show();
}