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.
assistant
#Here is Python program to determine whether Arkady held the round right after creating this problem, or not.

import sys
readline = sys.stdin.readline
class Segtree:
    def __init__(self, A, intv, initialize = True, segf = max):
        self.N = len(A)
        self.N0 = 2**(self.N-1).bit_length()
        self.intv = intv
        self.segf = segf
        if initialize:
            self.data = [intv]*self.N0 + A + [intv]*(self.N0 - self.N)
            for i in range(self.N0-1, 0, -1):
                self.data[i] = self.segf(self.data[2*i], self.data[2*i+1]) 
        else:
            self.data = [intv]*(2*self.N0)
        
    def add(self, k, x):
        k += self.N0
        self.data[k] += x
        while k > 0 :
            k = k >> 1
            self.data[k] = self.segf(self.data[2*k], self.data[2*k+1])
    
    def query(self, l, r):
        L, R = l+self.N0, r+self.N0
        s = self.intv
        while L < R:
            if R & 1:
                R -= 1
                s = self.segf(s, self.data[R])
            if L & 1:
                s = self.segf(s, self.data[L])
                L += 1
            L >>= 1
            R >>= 1
        return s
    
    def binsearch(self, l, r, check, reverse = False):
        L, R = l+self.N0, r+self.N0
        SL, SR = [], []
        while L < R:
            if R & 1:
                R -= 1
                SR.append(R)
            if L & 1:
                SL.append(L)
                L += 1
            L >>= 1
            R >>= 1
        
        if reverse:
            for idx in (SR + SL[::-1]):
                if check(self.data[idx]):
                    break
            else:
                return -1
            while idx < self.N0:
                if check(self.data[2*idx+1]):
                    idx = 2*idx + 1
                else:
                    idx = 2*idx
            return idx - self.N0
        else:
            for idx in (SL + SR[::-1]):
                if check(self.data[idx]):
                    break
            else:
                return -1
            while idx < self.N0:
                if check(self.data[2*idx]):
                    idx =...