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 
4
3 4
1 2
3 4
1 2
 4
2 4
1 3
4 4
1 1
 10
6 6
1 2
2 2
6 7
8 8
3 5
5 5
4 4
8 9
10 10
 4
1 4
1 4
2 4
1 2
 8
3 4
7 8
2 6
1 3
6 7
5 8
4 6
8 8
 7
7 7
2 3
2 3
5 7
5 6
4 4
1 2
 15
2 2
10 11
2 5
6 7
7 12
9 9
2 3
5 14
1 3
10 10
15 15
12 12
5 6
6 6
14 15
 5
1 3
4 4
5 5
1 4
1 5
 4
1 4
2 3
2 3
4 4
 4
1 4
1 4
1 2
1 4
 5
1 3
1 3
1 5
3 4
1 5
 3
1 3
2 2
1 3
 5
4 4
1 4
3 4
1 3
3 5
 3
1 3
1 2
2 3
 4
2 4
1 3
2 3
1 2
 5
4 5
4 5
1 3
1 2
2 4
 6
2 6
4 6
6 6
2 2
3 3
1 3
 3
2 3
1 2
1 3
 6
4 6
1 3
4 6
4 4
1 4
1 2
 7
3 6
6 6
5 7
2 4
2 3
1 2
3 5
 8
4 6
8 8
1 4
4 5
6 8
1 2
3 6
1 4
 6
2 4
2 6
4 5
2 5
6 6
1 4
 8
4 4
7 7
3 5
5 7
1 8
4 5
1 5
2 2
 4
3 4
1 3
2 3
1 2
 10
9 9
3 4
1 1
5 5
7 7
3 7
6 10
3 3
10 10
1 3
 4
1 4
4 4
2 3
1 1
 1
1 1
 8
1 5
6 8
1 3
4 8
1 1
3 3
6 6
8 8
 5
3 3
4 4
2 2
1 1
5 5
 4
1 1
2 3
2 3
1 4
 4
3 4
1 4
2 3
1 2
 2
1 2
1 2
 8
3 3
3 5
2 3
1 1
5 7
7 7
5 5
3 8
 5
2 2
3 3
1 2
1 5
4 4
 6
1 3
2 4
3 4
2 3
5 5
6 6
 6
1 4
3 6
6 6
3 5
1 5
2 5
 5
4 4
2 3
4 5
1 4
2 4
 3
1 2
1 2
1 3
 4
1 4
2 2
3 3
1 4
 5
2 5
1 1
2 3
3 4
5 5
 4
2 4
1 4
3 4
1 3
 6
1 1
4 6
3 6
1 3
3 4
3 6
 5
2 3
3 4
4 5
1 1
2 3
 5
2 3
1 5
2 4
3 5
4 4
 20
12 14
13 13
18 18
3 3
7 8
11 11
10 13
2 2
9 10
3 5
7 8
15 17
14 15
9 9
20 20
15 16
6 6
19 19
1 1
5 5
 5
1 5
2 2
3 3
4 4
1 5
 7
5 7
1 5
1 2
1 5
5 6
3 5
7 7

NO
3 1 4 2
3 2 4 1
 NO
3 2 4 1 
2 3 4 1  YES
6 1 2 7 8 3 5 4 9 10  NO
2 3 4 1 
1 3 4 2 
 YES
3 7 2 1 6 5 4 8  NO
7 2 3 6 5 4 1 
7 3 2 6 5 4 1 
 YES
2 11 4 7 8 9 3 13 1 10 15 12 5 6 14  NO
1 4 5 2 3 
2 4 5 1 3  NO
1 2 3 4 
1 3 2 4 
 NO
2 3 1 4 
1 3 2 4 
 NO
1 2 4 3 5 
2 1 4 3 5 
 NO
1 2 3 
3 2 1 
 NO
4 2 3 1 5 
4 1 3 2 5  NO
2 1 3
1 2 3
 NO
4 2 3 1
4 1 3 2
 NO
4 5 2 1 3
4 5 1 2 3
 NO
4 5 6 2 3 1 
5 4 6 2 3 1 
 NO
2 1 3 
3 1 2 
 NO
5 2 6 4 3 1 
5 1 6 4 3 2 
 NO
5 6 7 3 2 1 4 
5 6 7 2 3 1 4  NO
5 8 2 4 7 1 6 3
5 8 1 4 7 2 6 3
 NO
2 5 4 3 6 1 
3 5 4 2 6 1  YES
4 7 3 6 8 5 1 2  NO
4 2 3 1
4 1 3 2
 YES
9 4 1 5 7 6 8 3 10 2  NO
3 4 2 1 
2 4 3 1  YES
1  NO
4 7 2 5 1 3 6 8 
5 7 2 4 1 3 6 8  YES
3 4 2 1 5  NO
1 2 3 4 
1 3 2 4 
 NO
3 4 2 1 
4 3 2 1 
 NO
1 2 
2 1 
 YES
3 4 2 1 6 7 5 8  YES
2 3 1 5 4  NO
1 3 4 2 5 6
1 2 4 3 5 6
 NO
1 5 6 3 2 4 
2 5 6 3 1 4 
 NO
4 2 5 1 3 
4 3 5 1 2  NO
1 2 3 
2 1 3 
 NO
1 2 3 4 
4 2 3 1 
 NO
4 1 2 3 5 
3 1 2 4 5 
 NO
2 3 4 1
2 1 4 3
 NO
1 4 5 2 3 6 
1 5 4 2 3 6 
 NO
2 4 5 1 3 
3 4 5 1 2 
 NO
2 1 3 5 4 
3 1 2 5 4  NO
14 13 18 3 7 11 12 2 10 4 8 17 15 9 20 16 6 19 1 5 
14 13 18 3 8 11 12 2 10 4 7 17 15 9 20 16 6 19 1 5 
 NO
1 2 3 4 5 
5 2 3 4 1 
 NO
6 2 1 3 5 4 7
6 1 2 3 5 4 7

#include <bits/stdc++.h>
using namespace std;
template <typename A, typename B>
ostream& operator<<(ostream& out, const pair<A, B>& a) {
  out << "(" << a.first << "," << a.second << ")";
  return out;
}
template <typename T, size_t N>
ostream& operator<<(ostream& out, const array<T, N>& a) {
  out << "[";
  bool first = true;
  for (auto& v : a) {
    out << (first ? "" : ", ");
    out << v;
    first = 0;
  }
  out << "]";
  return out;
}
template <typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
  out << "[";
  bool first = true;
  for (auto& v : a) {
    out << (first ? "" : ", ");
    out << v;
    first = 0;
  }
  out << "]";
  return out;
}
template <typename T, class Cmp>
ostream& operator<<(ostream& out, const set<T, Cmp>& a) {
  out << "{";
  bool first = true;
  for (auto& v : a) {
    out << (first ? "" : ", ");
    out << v;
    first = 0;
  }
  out << "}";
  return out;
}
template <typename U, typename T, class Cmp>
ostream& operator<<(ostream& out, const map<U, T, Cmp>& a) {
  out << "{";
  bool first = true;
  for (auto& p : a) {
    out << (first ? "" : ", ");
    out << p.first << ":" << p.second;
    first = 0;
  }
  out << "}";
  return out;
}
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cerr << name << ": " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << ": " << arg1 << " |";
  __f(comma + 1, args...);
}
const int INF = 1 << 29;
const int MOD = 1e9 + 7;
mt19937 mrand(random_device{}());
int rnd(int x) { return mrand() % x; }
struct fast_ios {
  fast_ios() {
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(10);
  };
} fast_ios_;
template <typename T, size_t N>
struct fenwick {
  T c[N + 1];
  void clear() { fill(c, c + N + 1, 0); }
  void add(int x, T y) {
    for (; x <= N; x += x & -x) {
      c[x] += y;
    }
  }
  T query(int x) {
    T ret = 0;
    for (; x; x -= x & -x) {
      ret += c[x];
    }
    return ret;
  }
};
const int N = 2e5 + 10;
fenwick<int, N> C;
struct Node {
  int L, R, idx;
};
int main() {
  int n;
  scanf("%d", &n);
  vector<Node> a(n);
  for (int i = 0; i < n; ++i) {
    scanf("%d%d", &a[i].L, &a[i].R);
    a[i].idx = i;
  }
  sort(a.begin(), a.end(), [](const Node& u, const Node& v) {
    if (u.R != v.R) return u.R < v.R;
    return u.L < v.L;
  });
  pair<int, int> D = {-1, -1};
  for (int i = 1; i < n; ++i) {
    if (a[i].L == a[i - 1].L && a[i].R == a[i - 1].R) {
      D = {a[i].idx, a[i - 1].idx};
      break;
    }
  }
  vector<int> A(n), B(n);
  for (int i = 0; i < n; ++i) {
    auto [L, R, idx] = a[i];
    int low = L, high = R;
    int base = C.query(L - 1);
    while (low != high) {
      int mid = (low + high) / 2;
      if (mid - L + 1 + base == C.query(mid)) {
        low = mid + 1;
      } else {
        high = mid;
      }
    }
    C.add(high, 1);
    A[idx] = high;
  }
  42;
  sort(a.begin(), a.end(), [](const Node& u, const Node& v) {
    if (u.L != v.L) return u.L > v.L;
    return u.R < v.R;
  });
  C.clear();
  for (int i = 0; i < n; ++i) {
    auto [L, R, idx] = a[i];
    int low = L, high = R;
    int base = C.query(n - R);
    while (low != high) {
      int mid = (low + high + 1) / 2;
      if (R - mid + 1 + base == C.query(n + 1 - mid)) {
        high = mid - 1;
      } else {
        low = mid;
      }
    }
    C.add(n + 1 - high, 1);
    B[idx] = high;
  }
  42;
  if (A == B) {
    if (D == pair<int, int>{-1, -1}) {
      puts("YES");
      for (int i = 0; i < n; ++i) {
        printf("%d%c", A[i], " \n"[i + 1 == n]);
      }
    } else {
      puts("NO");
      for (int i = 0; i < n; ++i) {
        printf("%d%c", A[i], " \n"[i + 1 == n]);
      }
      swap(A[D.first], A[D.second]);
      for (int i = 0; i < n; ++i) {
        printf("%d%c", A[i], " \n"[i + 1 == n]);
      }
    }
  } else {
    puts("NO");
    for (int i = 0; i < n; ++i) {
      printf("%d%c", A[i], " \n"[i + 1 == n]);
    }
    for (int i = 0; i < n; ++i) {
      printf("%d%c", B[i], " \n"[i + 1 == n]);
    }
  }
  return 0;
}
