Problem:
Score : 1400 points

Problem Statement

You are given P, a permutation of (1,\ 2,\ ...\ N).

A string S of length N consisting of 0 and 1 is a good string when it meets the following criterion:

The sequences X and Y are constructed as follows:

First, let X and Y be empty sequences.

For each i=1,\ 2,\ ...\ N, in this order, append P_i to the end of X if S_i= 0, and append it to the end of Y if S_i= 1.

If X and Y have the same number of high elements, S is a good string.
Here, the i-th element of a sequence is called high when that element is the largest among the elements from the 1-st to i-th element in the sequence.

Determine if there exists a good string. If it exists, find the lexicographically smallest such string.

Constraints

1 \leq N \leq 2 \times 10^5

1 \leq P_i \leq N

P_1,\ P_2,\ ...\ P_N are all distinct.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
P_1 P_2 ... P_N

Output

If a good string does not exist, print -1.
If it exists, print the lexicographically smallest such string.

Sample Input 1

6
3 1 4 6 2 5

Sample Output 1

001001

Let S= 001001. Then, X=(3,\ 1,\ 6,\ 2) and Y=(4,\ 5).
The high elements in X is the first and third elements, and the high elements in Y is the first and second elements.
As they have the same number of high elements, 001001 is a good string.
There is no good string that is lexicographically smaller than this, so the answer is 001001.

Sample Input 2

5
1 2 3 4 5

Sample Output 2

-1

Sample Input 3

7
1 3 2 5 6 4 7

Sample Output 3

0001101

Sample Input 4

30
1 2 6 3 5 7 9 8 11 12 10 13 16 23 15 18 14 24 22 26 19 21 28 17 4 27 29 25 20 30

Sample Output 4

000000000001100101010010011101

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

// Written by newbiechd
#include <algorithm>
#include <cassert>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <vector>

// Delete the debugging information!
#define debug(x) std::cerr << #x << " = " << x << std::endl;

const int BUFFER_SIZE = 1 << 25 | 1;
struct InputOutputStream {
  char ibuf[BUFFER_SIZE], *s, *oh;
  InputOutputStream() : s(ibuf) {
    ibuf[fread(ibuf, 1, BUFFER_SIZE, stdin)] = '\0';
  }
  template <typename T>
  inline InputOutputStream &operator>>(T &x) {
    while (!isdigit(*s)) ++s;
    for (x = 0; isdigit(*s); ++s)
      x = x * 10 + (*s ^ '0');
    return *this;
  }
}IO;

template <typename lhs, typename rhs>
inline lhs min(const lhs &x, const rhs &y) { return x < y ? x : y; }
template <typename lhs, typename rhs>
inline lhs max(const lhs &x, const rhs &y) { return x > y ? x : y; }

const int MAXN = 200003;
int n, p[MAXN], suf[MAXN];
bool flag[MAXN];

const int MAXT = 800003, inf = 0x3f3f3f3f;
#define mid ((l + r) >> 1)
#define lSon (k << 1)
#define rSon (k << 1 | 1)
struct SegmentTree {
  int mx[MAXT];
  inline void init() { memset(mx, -0x3f, sizeof mx); }
  inline void pushup(int k) { mx[k] = max(mx[lSon], mx[rSon]); }
  void modify(int k, int l, int r, int x, int v) {
    if (l == r) {
      mx[k] = v;
      return ;
    }
    x <= mid ? modify(lSon, l, mid, x, v) : modify(rSon, mid + 1, r, x, v);
    pushup(k);
  }
  int query(int k, int l, int r, int x, int y) {
    if (x <= l && r <= y)
      return mx[k];
    int o = -inf;
    if (x <= mid)
      o = max(o, query(lSon, l, mid, x, y));
    if (mid < y)
      o = max(o, query(rSon, mid + 1, r, x, y));
    return o;
  }
}tree[2];
#undef mid
#undef lSon
#undef rSon

int f[MAXN][2], arrA[MAXN], arrB[MAXN];
char ans[MAXN];

int cntA, cntB, mxA, mxB;
inline bool check(int i) {
  int temp = cntA + (p[i] > mxA) + suf[i + 1] - cntB;
  if (temp >= 0 && tree[temp & 1].query(1, 1, n, mxB, n) >= temp)
      return 1;
  temp = cntB + suf[i + 1] - cntA - (p[i] > mxA);
  if (temp >= 0 && tree[temp & 1].query(1, 1, n, max(mxA, p[i]), n) >= temp)
    return 1;
  return 0;
}

int main() {
  IO >> n;
  for (int i = 1, mx = 0; i <= n; ++i) {
    IO >> p[i];
    if (p[i] > mx)
      flag[i] = 1, mx = p[i];
  }
  tree[1].init();
  for (int i = n; i; --i) {
    suf[i] = suf[i + 1] + flag[i];
    for (int j = 0; j < 2; ++j)
      f[i][j] = tree[j ^ (!flag[i])].query(1, 1, n, p[i], n) + 1 + flag[i];
    for (int j = 0; j < 2; ++j)
      tree[j].modify(1, 1, n, p[i], f[i][j]);
  }
  
  for (int i = 1; i <= n; ++i) {
    tree[0].modify(1, 1, n, p[i], 0);
    tree[1].modify(1, 1, n, p[i], -inf);
    if (check(i)) {
      ans[i] = '0';
      if (p[i] > mxA)
        mxA = p[i], ++cntA;
    }
    else {
      ans[i] = '1';
      if (p[i] > mxB)
        mxB = p[i], ++cntB;
    }
  }
  if (cntA == cntB)
    printf("%s\n", ans + 1);
  else
    printf("-1\n");
  return 0;
}