Problem: Two integer sequences existed initially — one of them was strictly increasing, and the other one — strictly decreasing.

Strictly increasing sequence is a sequence of integers [x_1 < x_2 < ... < x_k]. And strictly decreasing sequence is a sequence of integers [y_1 > y_2 > ... > y_l]. Note that the empty sequence and the sequence consisting of one element can be considered as increasing or decreasing.

They were merged into one sequence a. After that sequence a got shuffled. For example, some of the possible resulting sequences a for an increasing sequence [1, 3, 4] and a decreasing sequence [10, 4, 2] are sequences [1, 2, 3, 4, 4, 10] or [4, 2, 1, 10, 4, 3].

This shuffled sequence a is given in the input.

Your task is to find any two suitable initial sequences. One of them should be strictly increasing and the other one — strictly decreasing. Note that the empty sequence and the sequence consisting of one element can be considered as increasing or decreasing.

If there is a contradiction in the input and it is impossible to split the given sequence a to increasing and decreasing sequences, print "NO".

Input

The first line of the input contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of elements in a.

The second line of the input contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 2 ⋅ 10^5), where a_i is the i-th element of a.

Output

If there is a contradiction in the input and it is impossible to split the given sequence a to increasing and decreasing sequences, print "NO" in the first line.

Otherwise print "YES" in the first line and any two suitable sequences. Note that the empty sequence and the sequence consisting of one element can be considered as increasing or decreasing.

In the second line print n_i — the number of elements in the strictly increasing sequence. n_i can be zero, in this case the increasing sequence is empty.

In the third line print n_i integers inc_1, inc_2, ..., inc_{n_i} in the increasing order of its values (inc_1 < inc_2 < ... < inc_{n_i}) — the strictly increasing sequence itself. You can keep this line empty if n_i = 0 (or just print the empty line).

In the fourth line print n_d — the number of elements in the strictly decreasing sequence. n_d can be zero, in this case the decreasing sequence is empty.

In the fifth line print n_d integers dec_1, dec_2, ..., dec_{n_d} in the decreasing order of its values (dec_1 > dec_2 > ... > dec_{n_d}) — the strictly decreasing sequence itself. You can keep this line empty if n_d = 0 (or just print the empty line).

n_i + n_d should be equal to n and the union of printed sequences should be a permutation of the given sequence (in case of "YES" answer).

Examples

Input


7
7 2 7 3 3 1 4


Output


YES
2
3 7 
5
7 4 3 2 1 


Input


5
4 3 1 5 3


Output


YES
1
3 
4
5 4 3 1 


Input


5
1 1 2 1 2


Output


NO


Input


5
0 1 2 3 4


Output


YES
0

5
4 3 2 1 0 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
template <typename T>
void MACRO_VAR_Scan(T& t) {
  std::cin >> t;
}
template <typename First, typename... Rest>
void MACRO_VAR_Scan(First& first, Rest&... rest) {
  std::cin >> first;
  MACRO_VAR_Scan(rest...);
}
template <typename T>
void MACRO_VEC_ROW_Init(int n, T& t) {
  t.resize(n);
}
template <typename First, typename... Rest>
void MACRO_VEC_ROW_Init(int n, First& first, Rest&... rest) {
  first.resize(n);
  MACRO_VEC_ROW_Init(n, rest...);
}
template <typename T>
void MACRO_VEC_ROW_Scan(int p, T& t) {
  std::cin >> t[p];
}
template <typename First, typename... Rest>
void MACRO_VEC_ROW_Scan(int p, First& first, Rest&... rest) {
  std::cin >> first[p];
  MACRO_VEC_ROW_Scan(p, rest...);
}
template <class T>
inline T CHMAX(T& a, const T b) {
  return a = (a < b) ? b : a;
}
template <class T>
inline T CHMIN(T& a, const T b) {
  return a = (a > b) ? b : a;
}
template <class T>
using V = std::vector<T>;
template <class T>
using VV = V<V<T>>;
template <typename S, typename T>
std::ostream& operator<<(std::ostream& os, std::pair<S, T> p) {
  os << "(" << p.first << ", " << p.second << ")";
  return os;
}
using ll = long long;
using ull = unsigned long long;
using ld = long double;
using PAIR = std::pair<ll, ll>;
using PAIRLL = std::pair<ll, ll>;
constexpr ll INFINT = (1 << 30) - 1;
constexpr ll INFINT_LIM = (1LL << 31) - 1;
constexpr ll INFLL = 1LL << 60;
constexpr ll INFLL_LIM = (1LL << 62) - 1 + (1LL << 62);
constexpr double EPS = 1e-10;
constexpr ll MOD = 1000000007;
constexpr double PI = 3.141592653589793238462643383279;
template <class T, size_t N>
void FILL(T (&a)[N], const T& val) {
  for (auto& x : a) x = val;
}
template <class ARY, size_t N, size_t M, class T>
void FILL(ARY (&a)[N][M], const T& val) {
  for (auto& b : a) FILL(b, val);
}
template <class T>
void FILL(std::vector<T>& a, const T& val) {
  for (auto& x : a) x = val;
}
template <class ARY, class T>
void FILL(std::vector<std::vector<ARY>>& a, const T& val) {
  for (auto& b : a) FILL(b, val);
}
ll dp[200005][2];
ll pre[200005][2];
signed main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(0);
  ;
  ll n;
  MACRO_VAR_Scan(n);
  ;
  std::vector<ll> a(n);
  for (auto& i : a) std::cin >> i;
  ;
  FILL(dp, -1LL);
  dp[0][0] = INFINT;
  dp[0][1] = -INFINT;
  FILL(pre, -1LL);
  for (ll i = (1); i < (n); ++i) {
    {
      if (a[i - 1] < a[i]) {
        if (dp[i - 1][0] != -1) {
          dp[i][0] = dp[i - 1][0];
          pre[i][0] = 0;
        }
      }
      if (dp[i - 1][1] != -1 && dp[i - 1][1] < a[i]) {
        if (dp[i][0] == -1 || dp[i][0] < a[i - 1]) {
          dp[i][0] = a[i - 1];
          pre[i][0] = 1;
        }
      }
    }
    {
      if (a[i - 1] > a[i]) {
        if (dp[i - 1][1] != -1) {
          dp[i][1] = dp[i - 1][1];
          pre[i][1] = 1;
        }
      }
      if (dp[i - 1][0] != -1 && dp[i - 1][0] > a[i]) {
        if (dp[i][1] == -1 || dp[i][1] > a[i - 1]) {
          dp[i][1] = a[i - 1];
          pre[i][1] = 0;
        }
      }
    }
  }
  if (dp[n - 1][0] == -1 && dp[n - 1][1] == -1) {
    std::cout << ("NO");
    std::cout << "\n";
    ;
    return 0;
  }
  V<ll> ans(n);
  ll now = 0;
  if (dp[n - 1][0] == -1) now = 1;
  ans[n - 1] = now;
  for (ll i = (n)-1; i >= (1); --i) {
    now = pre[i][now];
    ans[i - 1] = now;
  }
  std::cout << ("YES");
  std::cout << "\n";
  ;
  for (ll i = 0; i < ll(n); ++i) {
    std::cout << (ans[i]);
    std::cout << (i + 1 == n ? '\n' : ' ');
    ;
  }
  return 0;
}