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.
Solve the task in PYTHON3.
 Can you provide a Python code to determine whether Arkady held the round right after creating this problem, or not?
assistant
#Here is Python code to determine whether Arkady held the round right after creating this problem, or not.

def chPathToThis(f=None):
    import os
    f = __file__ if f is None else f
    os.chdir(os.path.dirname(f))

fle = None

def input():
    global fle
    def inner():
        if fle is None:
            from sys import stdin
            for line in stdin:
                yield line
            raise EOFError
        else:
            for line in fle:
                yield line.replace('\r', '').replace('\n', '')
            raise EOFError
    o = inner()
    return next(o)

import math

class SegmentTree:
    
    def __init__(self, A):
        self.N = len(A)
        self.st = [0] * (4 * self.N) # approximate the overall size of segment tree with array N
        self.build(1, 0, self.N - 1)
        
    def left(self, idx):
        return idx * 2

    def right(self, idx):
        return idx * 2 + 1

    def build(self, idx, l, r):
        if l == r:
            self.st[idx] = A[l]
        else:
            mid = (l + r) // 2
            self.build(self.left(idx), l, mid)
            self.build(self.right(idx), mid + 1, r)
            self.st[idx] = max(self.st[self.left(idx)] , self.st[self.right(idx)])
    
    def update(self, a, b, val):
        return self.update_recursive(1, 0, self.N - 1, a - 1, b - 1, val)
    
    def update_recursive(self, idx, l, r, a, b, val): # update(1, 1, N, a, b, v) for update val v to [a,b]
        if r < a or l > b:
            return True
        if l == r :
            self.st[idx] = val
            return True
        mid = (l+r)//2
        self.update_recursive(self.left(idx), l, mid, a, b, val)
        self.update_recursive(self.right(idx), mid+1, r, a, b, val)
        self.st[idx] = max(self.st[self.left(idx)] , self.st[self.right(idx)])
        return True

    def query(self, a, b):
        return self.query_recursive(1, 0, self.N - 1, a - 1, b - 1)

    def query_recursive(self, idx, l, r, a, b): #query(1, 1, N, a, b) for query max of [a,b]
        if r < a or l > b:
            return -math.inf
        if l >= a and r <= b:
            return self.st[idx]
       ...