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;
struct interv {
  int x, y, p;
  bool operator<(const interv &aux) const {
    if (x != aux.x) return x < aux.x;
    return y < aux.y;
  }
};
interv a[200005];
int n;
int which[200005], ans[200005], aux[200005];
set<pair<int, int> > s;
int Arb[800020], where[800020];
void update(int pos, int val, int st = 1, int dr = n, int nod = 1) {
  if (st == dr) {
    Arb[nod] = val;
    if (val == 1)
      where[nod] = pos;
    else
      where[nod] = 0;
    return;
  }
  int mij = (st + dr) / 2;
  if (pos <= mij) update(pos, val, st, mij, nod * 2);
  if (mij + 1 <= pos) update(pos, val, mij + 1, dr, nod * 2 + 1);
  Arb[nod] = Arb[nod * 2] + Arb[nod * 2 + 1];
  where[nod] = max(where[nod * 2], where[nod * 2 + 1]);
}
int query(int x, int y, int st = 1, int dr = n, int nod = 1) {
  if (x <= st && dr <= y) return where[nod];
  int mij = (st + dr) / 2;
  int ans = 0;
  if (x <= mij) ans = max(ans, query(x, y, st, mij, nod * 2));
  if (mij + 1 <= y) ans = max(ans, query(x, y, mij + 1, dr, nod * 2 + 1));
  return ans;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d%d", &a[i].x, &a[i].y);
    a[i].p = i;
  }
  sort(a + 1, a + n + 1);
  int j = 1;
  for (int val = 1; val <= n; ++val) {
    while (j <= n && a[j].x <= val) {
      s.insert({a[j].y, j});
      ++j;
    }
    which[val] = s.begin()->second;
    aux[s.begin()->second] = val;
    s.erase(s.begin());
  }
  for (int val = 1; val <= n; ++val) ans[a[which[val]].p] = val;
  j = 1;
  for (int val = 1; val <= n; ++val) {
    while (j <= n && a[j].x <= val) {
      update(aux[j], 1);
      s.insert({a[j].y, j});
      ++j;
    }
    int q = 0;
    int ind = which[val];
    if (a[ind].x < val) q = max(q, query(a[ind].x, val - 1));
    if (val < a[ind].y && q == 0) q = max(q, query(val + 1, a[ind].y));
    if (q != 0) {
      printf("NO\n");
      for (int i = 1; i <= n; ++i) printf("%d ", ans[i]);
      ans[a[ind].p] = q;
      ans[a[which[q]].p] = val;
      printf("\n");
      for (int i = 1; i <= n; ++i) printf("%d ", ans[i]);
      return 0;
    }
    while (!s.empty() && s.begin()->first == val) {
      int j = s.begin()->second;
      update(aux[j], 0);
      s.erase(s.begin());
    }
  }
  printf("YES\n");
  for (int i = 1; i <= n; ++i) printf("%d ", ans[i]);
  return 0;
}
