Two integer sequences existed initially, one of them was strictly increasing, and another 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.

Elements of increasing sequence were inserted between elements of the decreasing one (and, possibly, before its first element and after its last element) without changing the order. For example, sequences [1, 3, 4] and [10, 4, 2] can produce the following resulting sequences: [10, 1, 3, 4, 2, 4], [1, 3, 4, 10, 4, 2]. The following sequence cannot be the result of these insertions: [1, 10, 4, 4, 3, 2] because the order of elements in the increasing sequence was changed.

Let the obtained sequence be a. This 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 another 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 into one increasing sequence and one decreasing sequence, 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 into one increasing sequence and one decreasing sequence, print "NO" in the first line.

Otherwise print "YES" in the first line. In the second line, print a sequence of n integers res_1, res_2, ..., res_n, where res_i should be either 0 or 1 for each i from 1 to n. The i-th element of this sequence should be 0 if the i-th element of a belongs to the increasing sequence, and 1 otherwise. Note that the empty sequence and the sequence consisting of one element can be considered as increasing or decreasing.

Examples

Input


9
5 1 3 6 8 2 9 0 10


Output


YES
1 0 0 0 0 1 0 1 0 


Input


5
1 2 4 0 2


Output


NO
#include <bits/stdc++.h>
#pragma GCC target("avx")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using std::cin;
using std::cout;
using std::lower_bound;
using std::string;
using std::upper_bound;
using std::vector;
using vi = vector<long long>;
using vii = vector<vi>;
using pii = std::pair<long long, long long>;
constexpr long long MOD = 1e9 + 7;
constexpr long long MAX = 3e6;
constexpr long long inf = (1ll << 60);
template <class T>
class prique : public std::priority_queue<T, std::vector<T>, std::greater<T>> {
};
template <typename T>
struct Segment_tree {
  long long N;
  T mem;
  vector<T> node;
  Segment_tree(vector<T> &X, T m) : mem(m) {
    long long sz = X.size();
    N = 1;
    while (N < sz) N *= 2;
    node.resize(2 * N - 1, mem);
    for (long long i = (long long)(0); i < (long long)(sz); i++)
      node[N - 1 + i] = X[i];
    for (long long i = (long long)(N - 2); (long long)(0) <= i; i--) {
      node[i] = Compare(node[i * 2 + 1], node[i * 2 + 2]);
    }
  }
  T Compare(T &A, T &B) { return std::max(A, B); }
  void update(long long X, T val) {
    X += N - 1;
    node[X] = val;
    while (X > 0) {
      X = (X - 1) / 2;
      node[X] = Compare(node[X * 2 + 1], node[X * 2 + 2]);
    }
  }
  T Query(long long a, long long b, long long now, long long l, long long r) {
    if (r < 0) r = N;
    if (r <= a || b <= l) return mem;
    if (a <= l && r <= b) return node[now];
    auto vl = Query(a, b, now * 2 + 1, l, (l + r) / 2),
         vr = Query(a, b, now * 2 + 2, (l + r) / 2, r);
    return Compare(vl, vr);
  }
};
template <typename T>
struct lazy_Segment_tree {
  int N;
  vector<T> node, lazy;
  T INF;
  lazy_Segment_tree(vector<T> X, T Y) : INF(Y) {
    N = 1;
    while (X.size() > N) N *= 2;
    node.resize(2 * N - 1, Y);
    lazy.resize(2 * N - 1);
    for (long long i = (long long)(0); i < (long long)(X.size()); i++)
      node[i + N - 1] = X[i];
    for (long long i = (long long)(N - 2); (long long)(0) <= i; i--)
      node[i] = compare(node[i * 2 + 1], node[i * 2 + 2]);
  }
  T compare(T X, T Y) { return std::max(X, Y); }
  T plus(T X, int l, int r) { return X; }
  void eval(int now, int l, int r) {
    if (lazy[now] == 0) return;
    node[now] += lazy[now];
    if (r - l > 1) {
      lazy[now * 2 + 1] += plus(lazy[now], l, r);
      lazy[now * 2 + 2] += plus(lazy[now], l, r);
    }
    lazy[now] = 0;
  }
  void update(int a, int b, T add, int now = 0, int l = 0, int r = -1) {
    if (r < 0) r = N;
    eval(now, l, r);
    if (b <= l || r <= a) return;
    if (a <= l && r <= b) {
      lazy[now] += add;
      eval(now, l, r);
    } else {
      update(a, b, add, now * 2 + 1, l, (r + l) / 2);
      update(a, b, add, now * 2 + 2, (r + l) / 2, r);
      node[now] = compare(node[now * 2 + 1], node[now * 2 + 2]);
    }
  }
  T Query(int a, int b, int now = 0, int l = 0, int r = -1) {
    if (r < 0) r = N;
    eval(now, l, r);
    if (b <= l || r <= a) return INF;
    if (a <= l && r <= b) return node[now];
    return compare(Query(a, b, now * 2 + 1, l, (r + l) / 2),
                   Query(a, b, now * 2 + 2, (r + l) / 2, r));
  }
};
struct Tree {
  int N;
  vii dp;
  vi dist;
  Tree(vii edge) {
    N = edge.size();
    dp.resize(N);
    dist.resize(N, -1);
    for (int i = 0; i < N; i++) dp[i].resize(30);
    dist[0] = dp[0][0] = 0;
    std::queue<int> que;
    que.push(0);
    while (!que.empty()) {
      int now = que.front();
      que.pop();
      for (int i = 0; i < edge[now].size(); i++) {
        int next = edge[now][i];
        if (dist[next] == -1) {
          dist[next] = dist[now] + 1;
          que.push(next);
          dp[next][0] = now;
        }
      }
    }
    for (int i = 1; i < 30; i++) {
      for (int j = 0; j < N; j++) dp[j][i] = dp[dp[j][i - 1]][i - 1];
    }
  }
  int LCA(int X, int Y) {
    if (dist[X] < dist[Y]) std::swap(X, Y);
    {
      int Z = dist[X] - dist[Y];
      for (int i = 0; i < 30; i++) {
        if (Z & (1 << i)) {
          X = dp[X][i];
        }
      }
    }
    if (X == Y) return X;
    for (int i = 29; i >= 0; i--) {
      if (dp[X][i] != dp[Y][i]) {
        X = dp[X][i];
        Y = dp[Y][i];
      }
    }
    return dp[X][0];
  }
};
struct Binary_indexed_tree {
  int N;
  vi bit;
  Binary_indexed_tree(int n) : N(n) { bit.resize(N + 1, 0); }
  void add(int x, long long a) {
    for (x; x <= N; x += (x & -x)) bit[x] += a;
  }
  long long sum(int x) {
    long long ret = 0;
    for (x; x > 0; x -= (x & -x)) ret += bit[x];
    return ret;
  }
  long long lower_bound(long long X) {
    if (sum(N) < X) return -1;
    long long ret = 0, memo = 1, sum = 0;
    while (memo * 2 <= N) memo *= 2;
    while (memo > 0) {
      if (memo + ret <= N && sum + bit[memo + ret] < X) {
        sum += bit[memo + ret];
        ret += memo;
      }
      memo /= 2;
    }
    return ret + 1;
  }
};
struct Union_Find {
  long long N;
  vi par;
  vi siz;
  Union_Find(int n) : N(n) {
    par.resize(N);
    siz.resize(N, 1);
    for (long long i = (long long)(0); i < (long long)(N); i++) par[i] = i;
  }
  long long root(long long X) {
    if (par[X] == X) return X;
    return par[X] = root(par[X]);
  }
  bool same(long long X, long long Y) { return root(X) == root(Y); }
  void unite(long long X, long long Y) {
    X = root(X);
    Y = root(Y);
    if (X == Y) return;
    par[X] = Y;
    siz[Y] += siz[X];
    siz[X] = 0;
  }
  long long size(long long X) { return siz[root(X)]; }
};
long long modpow(long long a, long long n, long long mod) {
  long long res = 1;
  while (n > 0) {
    if (n & 1) res = res * a % mod;
    a = a * a % mod;
    n >>= 1;
  }
  return res;
}
vi fac, finv, inv;
void COMinit() {
  fac.resize(MAX);
  finv.resize(MAX);
  inv.resize(MAX);
  fac[0] = fac[1] = 1;
  finv[0] = finv[1] = 1;
  inv[1] = 1;
  for (int i = 2; i < MAX; i++) {
    fac[i] = fac[i - 1] * i % MOD;
    inv[i] = MOD - inv[MOD % i] * (MOD / i) % MOD;
    finv[i] = finv[i - 1] * inv[i] % MOD;
  }
}
long long COM(long long n, long long r) {
  if (n < r || n < 0 || r < 0) return 0;
  return fac[n] * finv[r] % MOD * finv[n - r] % MOD;
}
void comp(vi &A) {
  std::map<long long, long long> memo;
  for (long long i = (long long)(0); i < (long long)(A.size()); i++)
    memo[A[i]] = 0;
  long long cnt = 1;
  for (auto &p : memo) p.second = cnt++;
  for (long long i = (long long)(0); i < (long long)(A.size()); i++)
    A[i] = memo[A[i]];
}
void dec(std::map<long long, long long> &mem, long long X) {
  mem[X]--;
  if (mem[X] == 0) mem.erase(X);
}
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(nullptr);
  long long N;
  cin >> N;
  vi A(N);
  for (long long i = (long long)(0); i < (long long)(N); i++) cin >> A[i];
  vii dp(N + 1, vi(2));
  dp[0][0] = inf;
  dp[0][1] = -inf;
  vii p(N, vi(2, -1));
  for (long long i = (long long)(0); i < (long long)(N); i++) {
    dp[i + 1][0] = -inf;
    dp[i + 1][1] = inf;
    if (i == 0 || A[i - 1] < A[i]) {
      dp[i + 1][0] = dp[i][0];
      p[i][0] = 0;
    }
    if (i == 0 || A[i - 1] > A[i]) {
      dp[i + 1][1] = dp[i][1];
      p[i][1] = 1;
    }
    if (i && dp[i][0] > A[i]) {
      if (dp[i + 1][1] > A[i - 1]) p[i][1] = 0;
      dp[i + 1][1] = std::min(dp[i + 1][1], A[i - 1]);
    }
    if (i && dp[i][1] < A[i]) {
      if (dp[i + 1][0] < A[i - 1]) p[i][0] = 1;
      dp[i + 1][0] = std::max(dp[i + 1][0], A[i - 1]);
    }
    if (i == 0) {
      dp[i + 1][0] = inf;
      dp[i + 1][1] = -inf;
    }
    if (dp[i + 1][0] == -inf && dp[i + 1][1] == inf) {
      cout << "NO"
           << "\n";
      return 0;
    }
  }
  cout << "YES"
       << "\n";
  vi ans(N);
  if (dp[N][1] == inf)
    ans[N - 1] = 0;
  else
    ans[N - 1] = 1;
  for (long long i = (long long)(N - 2); (long long)(0) <= i; i--) {
    ans[i] = p[i + 1][ans[i + 1]];
  }
  for (long long i = (long long)(0); i < (long long)(N); i++)
    cout << ans[i] << " ";
  cout << "\n";
}
