human
Arkady coordinates rounds on some not really famous competitive programming platform. Each round features n problems of distinct difficulty, the difficulties are numbered from 1 to n.

To hold a round Arkady needs n new (not used previously) problems, one for each difficulty. As for now, Arkady creates all the problems himself, but unfortunately, he can't just create a problem of a desired difficulty. Instead, when he creates a problem, he evaluates its difficulty from 1 to n and puts it into the problems pool.

At each moment when Arkady can choose a set of n new problems of distinct difficulties from the pool, he holds a round with these problems and removes them from the pool. Arkady always creates one problem at a time, so if he can hold a round after creating a problem, he immediately does it.

You are given a sequence of problems' difficulties in the order Arkady created them. For each problem, determine whether Arkady held the round right after creating this problem, or not. Initially the problems pool is empty.

Input

The first line contains two integers n and m (1 ≤ n, m ≤ 10^5) — the number of difficulty levels and the number of problems Arkady created.

The second line contains m integers a_1, a_2, …, a_m (1 ≤ a_i ≤ n) — the problems' difficulties in the order Arkady created them.

Output

Print a line containing m digits. The i-th digit should be 1 if Arkady held the round after creation of the i-th problem, and 0 otherwise.

Examples

Input

3 11
2 3 1 2 2 2 3 2 2 3 1

Output

00100000001

Input

4 8
4 1 3 3 2 3 3 3

Output

00001000

Note

In the first example Arkady held the round after the first three problems, because they are of distinct difficulties, and then only after the last problem.
The above is tricky. Write me a correct solution in PYTHON3.
assistant
#Here is Python program to determine whether Arkady held the round right after creating this problem, or not.

''' Thruth can only be found at one place - THE CODE '''

''' Copyright 2018, SATYAM KUMAR'''
from sys import stdin, stdout
import cProfile, math
from collections import Counter
from bisect import bisect_left,bisect,bisect_right
import itertools
from copy import deepcopy
from fractions import Fraction
import sys, threading
sys.setrecursionlimit(10**6) # max depth of recursion
threading.stack_size(2**27)  # new thread will get stack of such size

printHeap = str()
memory_constrained = False
P = 10**9+7
import sys
sys.setrecursionlimit(10000000)

class Operation:
    def __init__(self, name, function, function_on_equal, neutral_value=0):
        self.name = name
        self.f = function
        self.f_on_equal = function_on_equal
def add_multiple(x, count):
    return x * count
def min_multiple(x, count):
    return x
def max_multiple(x, count):
    return x
sum_operation = Operation("sum", sum, add_multiple, 0)
min_operation = Operation("min", min, min_multiple, 1e9)
max_operation = Operation("max", max, max_multiple, -1e9)
class SegmentTree:
    def __init__(self,
                 array,
                 operations=[sum_operation, min_operation, max_operation]):
        self.array = array
        if type(operations) != list:
            raise TypeError("operations must be a list")
        self.operations = {}
        for op in operations:
            self.operations[op.name] = op
        self.root = SegmentTreeNode(0, len(array) - 1, self)
    def query(self, start, end, operation_name):
        if self.operations.get(operation_name) == None:
            raise Exception("This operation is not available")
        return self.root._query(start, end, self.operations[operation_name])
    def summary(self):
        return self.root.values
    def update(self, position, value):
        self.root._update(position, value)
    def update_range(self, start, end, value):
        self.root._update_range(start, end, value)
    def __repr__(self):
        return self.root.__repr__()
class SegmentTreeNode:
    def...