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>
const int N = 2e5 + 10;
const int inf = 0x3f3f3f3f;
int n, m, i, j, k;
int l[N], r[N];
int a[N], rev[N], pos[N], senq[N];
struct Segment_Tree1 {
  int mn[N << 2];
  inline void push_up(int u) { mn[u] = std::min(mn[u << 1], mn[u << 1 | 1]); }
  void build(int l, int r, int u) {
    if (l == r) {
      mn[u] = l;
      return;
    }
    int mid = (l + r) >> 1;
    build(l, mid, u << 1);
    build(mid + 1, r, u << 1 | 1);
    push_up(u);
  }
  int range_min(int ql, int qr, int l, int r, int u) {
    if (ql <= l && r <= qr) {
      return mn[u];
    }
    int mid = (l + r) >> 1, res = inf;
    if (ql <= mid) {
      res = std::min(res, range_min(ql, qr, l, mid, u << 1));
    }
    if (mid < qr) {
      res = std::min(res, range_min(ql, qr, mid + 1, r, u << 1 | 1));
    }
    return res;
  }
  void modify(int pos, int l, int r, int u, int val) {
    if (l == r) {
      mn[u] = val;
      return;
    }
    int mid = (l + r) >> 1;
    if (pos <= mid) {
      modify(pos, l, mid, u << 1, val);
    } else {
      modify(pos, mid + 1, r, u << 1 | 1, val);
    }
    push_up(u);
  }
} st1;
struct Segment_Tree2 {
  int mn[N << 2];
  inline int cmpl(int a, int b) { return l[rev[a]] < l[rev[b]] ? a : b; }
  inline void push_up(int u) { mn[u] = cmpl(mn[u << 1], mn[u << 1 | 1]); }
  void build(int l, int r, int u) {
    if (l == r) {
      mn[u] = l;
      return;
    }
    int mid = (l + r) >> 1;
    build(l, mid, u << 1);
    build(mid + 1, r, u << 1 | 1);
    push_up(u);
  }
  int range_min(int ql, int qr, int l, int r, int u) {
    if (ql <= l && r <= qr) {
      return mn[u];
    }
    int mid = (l + r) >> 1, res = 0;
    if (ql <= mid) {
      res = cmpl(res, range_min(ql, qr, l, mid, u << 1));
    }
    if (mid < qr) {
      res = cmpl(res, range_min(ql, qr, mid + 1, r, u << 1 | 1));
    }
    return res;
  }
} st2;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d %d", l + i, r + i);
    senq[i] = i;
  }
  std::sort(senq + 1, senq + n + 1, [](int a, int b) { return r[a] < r[b]; });
  l[0] = inf;
  st1.build(1, n, 1);
  for (int i = 1; i <= n; i++) {
    pos[senq[i]] = st1.range_min(l[senq[i]], r[senq[i]], 1, n, 1);
    rev[pos[senq[i]]] = senq[i];
    st1.modify(pos[senq[i]], 1, n, 1, n);
  }
  st2.build(1, n, 1);
  int x = -1, y = -1;
  bool flag = 0;
  for (int i = 1, cur; i <= n; i++) {
    if (l[cur = rev[st2.range_min(i + 1, r[rev[i]], 1, n, 1)]] <= i) {
      x = rev[i];
      y = cur;
      flag = 1;
      break;
    }
  }
  if (!flag) {
    puts("YES");
    for (int i = 1; i <= n; i++) {
      printf("%d%c", pos[i], " \n"[i == n]);
    }
  } else {
    puts("NO");
    for (int i = 1; i <= n; i++) {
      printf("%d%c", pos[i], " \n"[i == n]);
    }
    std::swap(pos[x], pos[y]);
    for (int i = 1; i <= n; i++) {
      printf("%d%c", pos[i], " \n"[i == n]);
    }
  }
  return 0;
}
