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.

Examples

Input

6
3 1 4 6 2 5


Output

001001


Input

5
1 2 3 4 5


Output

-1


Input

7
1 3 2 5 6 4 7


Output

0001101


Input

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


Output

000000000001100101010010011101
//Author:xht37
#include <bits/stdc++.h>

#define ui unsigned int
#define ll long long
#define ul unsigned ll
#define ld long double

#define pi pair <int, int>
#define fi first
#define se second
#define mp make_pair

#define ls (p << 1)
#define rs (ls | 1)
#define md ((t[p].l + t[p].r) >> 1)

#define vi vector <int>
#define pb push_back
#define pq priority_queue

#define dbg(x) cerr << #x" = " << x << endl
#define debug(...) fprintf(stderr, __VA_ARGS__)

#define fl(x) freopen(x".in", "r", stdin), freopen(x".out", "w", stdout)

using namespace std;

namespace io {
	const int SI = 1 << 21 | 1;
	char IB[SI], *IS, *IT, OB[SI], *OS = OB, *OT = OS + SI - 1, c, ch[100];
	int f, t;
	#define gc() (IS == IT ? (IT = (IS = IB) + fread(IB, 1, SI, stdin), IS == IT ? EOF : *IS++) : *IS++)
	inline void flush() {
		fwrite(OB, 1, OS - OB, stdout), OS = OB;
	}
	inline void pc(char x) {
		*OS++ = x;
		if (OS == OT) flush();
	}

	template <class I>
	inline void rd(I &x) {
		for (f = 1, c = gc(); c < '0' || c > '9'; c = gc()) if (c == '-') f = -1;
		for (x = 0; c >= '0' && c <= '9'; x = (x << 3) + (x << 1) + (c & 15), c = gc());
		x *= f;
	}
	template <class I>
	inline void rd(I &x, I &y) {
		rd(x), rd(y);
	}
	template <class I>
	inline void rd(I &x, I &y, I &z) {
		rd(x), rd(y), rd(z);
	}
	template <class I>
	inline void rda(I *a, int n) {
		for (int i = 1; i <= n; i++) rd(a[i]);
	}
	inline void rdc(char &c) {
		for (c = gc(); c < 33 || c > 126; c = gc());
	}
	inline void rds(char *s, int &n) {
		for (c = gc(); c < 33 || c > 126; c = gc());
		for (n = 0; c >= 33 && c <= 126; s[++n] = c, c = gc());
		s[n+1] = '\0';
	}
	inline void rds(string &s) {
		for (c = gc(); c < 33 || c > 126; c = gc());
		for (s.clear(); c >= 33 && c <= 126; s.pb(c), c = gc());
	}

	template <class I>
	inline void print(I x, char k = '\n') {
		if (!x) pc('0');
		if (x < 0) pc('-'), x = -x;
		while (x) ch[++t] = x % 10 + '0', x /= 10;
		while (t) pc(ch[t--]);
		pc(k);
	}
	template <class I>
	inline void print(I x, I y) {
		print(x, ' '), print(y);
	}
	template <class I>
	inline void print(I x, I y, I z) {
		print(x, ' '), print(y, ' '), print(z);
	}
	template <class I>
	inline void printa(I *a, int n) {
		for (int i = 1; i <= n; i++) print(a[i], " \n"[i==n]);
	}
	inline void printc(char c) {
		pc(c);
	}
	inline void prints(char *s, int n) {
		for (int i = 1; i <= n; i++) pc(s[i]);
		pc('\n');
	}
	inline void prints(string s) {
		int n = s.length();
		while (t < n) pc(s[t++]);
		pc('\n'), t = 0;
	}
	struct Flush {
		~Flush() {
			flush();
		}
	} flusher;
}
using io::rd;
using io::rda;
using io::rdc;
using io::rds;
using io::print;
using io::printa;
using io::printc;
using io::prints;

const int N = 2e5 + 7, inf = 1e9;
int n, p[N], c[N], cx, cy;
bool v[N];
char s[N];

struct SGT {
	struct T {
		int l, r, x;
	} t[N<<2|1];

	inline void update(int p) {
		t[p].x = max(t[ls].x, t[rs].x);
	}

	void build(int p, int l, int r, int x) {
		t[p].l = l, t[p].r = r;
		if (l == r) return t[p].x = x, void();
		build(ls, l, md, x), build(rs, md + 1, r, x);
		update(p);
	}

	void add(int p, int x, int k) {
		if (t[p].l == t[p].r) return t[p].x = k, void();
		if (x <= md) add(ls, x, k);
		if (x > md) add(rs, x, k);
		update(p);
	}

	int ask(int p, int l, int r) {
		if (t[p].l >= l && t[p].r <= r) return t[p].x;
		int ret = -inf;
		if (l <= md) ret = max(ret, ask(ls, l, r));
		if (r > md) ret = max(ret, ask(rs, l, r));
		return ret;
	}
} t[2];

inline bool pd(int x, int k) {
	return k < 0 ? 0 : t[k&1].ask(1, x, n) >= k;
}

int main() {
	rd(n), rda(p, n);
	for (int i = 1, x = 0; i <= n; i++)
		if (p[i] > x) x = p[i];
		else v[i] = 1;
	t[0].build(1, 1, n, 0), t[1].build(1, 1, n, -inf);
	for (int i = n; i; i--) {
		int x0 = t[0].ask(1, p[i], n), x1 = t[1].ask(1, p[i], n);
		t[0].add(1, p[i], (v[i] ? x1 + 1 : x0 + 2));
		t[1].add(1, p[i], (v[i] ? x0 + 1 : x1 + 2));
		c[i] = c[i+1] + !v[i];
	}
	for (int i = 1, x = 0, y = 0; i <= n; i++) {
		t[0].add(1, p[i], 0), t[1].add(1, p[i], -inf);
		if (pd(y, c[i+1] + cx - cy + (p[i] > x)) || pd(max(x, p[i]), c[i+1] + cy - cx - (p[i] > x)))
			s[i] = '0', cx += p[i] > x, x = max(x, p[i]);
		else s[i] = '1', cy += p[i] > y, y = max(y, p[i]); 
	}
	if (cx == cy) prints(s, n);
	else print(-1);
	return 0;
}