Problem: You are given an array a consisting of n integers. Initially all elements of a are either 0 or 1. You need to process q queries of two kinds:

  * 1 x : Assign to a_x the value 1 - a_x. 
  * 2 k : Print the k-th largest value of the array. 



As a reminder, k-th largest value of the array b is defined as following:

  * Sort the array in the non-increasing order, return k-th element from it. 



For example, the second largest element in array [0, 1, 0, 1] is 1, as after sorting in non-increasing order it becomes [1, 1, 0, 0], and the second element in this array is equal to 1.

Input

The first line contains two integers n and q (1 ≤ n, q ≤ 10^5) — the length of the given array and the number of queries.

The second line contains n integers a_1, a_2, a_3, ..., a_n (0 ≤ a_i ≤ 1) — elements of the initial array.

Each of the following q lines contains two integers. The first integer is t (1 ≤ t ≤ 2) — the type of query. 

  * If t = 1 the second integer is x (1 ≤ x ≤ n) — the position of the modified number. You have to assign to a_x the value 1 - a_x.
  * If t = 2 the second integer is k (1 ≤ k ≤ n) — you need to print the k-th largest value of the array.



It's guaranteed that there will be at least one query of the second type (satisfying t = 2).

Output

For each query of the second type, print a single integer — the answer to the query.

Example

Input


5 5
1 1 0 1 0
2 3
1 2
2 3
2 1
2 5


Output


1
0
1
0

Note

Initially a = [1, 1, 0, 1, 0].

The first operation is printing the third largest value, which is 1.

The second operation is assigning a_2 the value 0, a becomes [1, 0, 0, 1, 0].

The third operation is printing the third largest value, it is 0.

The fourth operation is printing the first largest value, it is 1.

The last operation is printing the fifth largest value, it is 0.

Solution (in PYTHON):

Solution:
from __future__ import division, print_function
from itertools import permutations 
import threading,bisect,math,heapq,sys
from collections import deque
# threading.stack_size(2**27)
# sys.setrecursionlimit(10**4)
from sys import stdin, stdout
i_m=9223372036854775807    
def cin():
    return map(int,sin().split())
def ain():                           #takes array as input
    return list(map(int,sin().split()))
def sin():
    return input()
def inin():
    return int(input()) 
prime=[]
def dfs(n,d,v):
    v[n]=1
    x=d[n]
    for i in x:
        if i not in v:
            dfs(i,d,v)
    return p 
def block(x): 
      
    v = []  
    while (x > 0): 
        v.append(int(x % 2)) 
        x = int(x / 2) 
    ans=[]
    for i in range(0, len(v)): 
        if (v[i] == 1): 
            ans.append(2**i)  
    return ans 

"""**************************MAIN*****************************"""
  
def main():
    n,k=cin()
    a=ain()
    p=0
    for i in a:
        if i==1:
            p+=1
    for _ in range(k):
        x,y=cin()
        # print(p,a)
        if x==2:
            if y<=p:
                print(1)
            else:
                print(0)
        else:
            if a[y-1]==1:
                a[y-1]=0
                p-=1
            else:
                a[y-1]=1
                p+=1    
"""***********************************************"""
def intersection(l,r,ll,rr):
    # print(l,r,ll,rr)
    if (ll > r or rr < l): 
            return 0
    else: 
        l = max(l, ll) 
        r = min(r, rr)
    return max(0,r-l+1) 
######## Python 2 and 3 footer by Pajenegod and c1729
fac=[]
def fact(n,mod):
    global fac
    fac.append(1)
    for i in range(1,n+1):
        fac.append((fac[i-1]*i)%mod)
    f=fac[:]
    return f
def nCr(n,r,mod):
    global fac
    x=fac[n]
    y=fac[n-r]
    z=fac[r]
    x=moddiv(x,y,mod)
    return moddiv(x,z,mod)
def moddiv(m,n,p):
    x=pow(n,p-2,p)
    return (m*x)%p
def GCD(x, y): 
    x=abs(x)
    y=abs(y)
    if(min(x,y)==0):
        return max(x,y)
    while(y): 
        x, y = y, x % y 
    return x 
def Divisors(n) : 
    l = []  
    ll=[]
    for i in range(1, int(math.sqrt(n) + 1)) :
        if (n % i == 0) : 
            if (n // i == i) : 
                l.append(i) 
            else : 
                l.append(i)
                ll.append(n//i)
    l.extend(ll[::-1])
    return l
def SieveOfEratosthenes(n): 
    global prime
    prime = [True for i in range(n+1)] 
    p = 2
    while (p * p <= n): 
        if (prime[p] == True): 
            for i in range(p * p, n+1, p): 
                prime[i] = False
        p += 1
    f=[]
    for p in range(2, n): 
        if prime[p]: 
            f.append(p)
    return f
def primeFactors(n): 
    a=[]
    while n % 2 == 0: 
        a.append(2) 
        n = n // 2 
    for i in range(3,int(math.sqrt(n))+1,2):  
        while n % i== 0: 
            a.append(i) 
            n = n // i  
    if n > 2: 
        a.append(n)
    return a
"""*******************************************************"""
py2 = round(0.5)
if py2:
    from future_builtins import ascii, filter, hex, map, oct, zip
    range = xrange
import os
from io import IOBase, BytesIO
BUFSIZE = 8192
class FastIO(BytesIO):
    newlines = 0
 
    def __init__(self, file):
        self._file = file
        self._fd = file.fileno()
        self.writable = "x" in file.mode or "w" in file.mode
        self.write = super(FastIO, self).write if self.writable else None
 
    def _fill(self):
        s = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
        self.seek((self.tell(), self.seek(0,2), super(FastIO, self).write(s))[0])
        return s
 
    def read(self):
        while self._fill(): pass
        return super(FastIO,self).read()
 
    def readline(self):
        while self.newlines == 0:
            s = self._fill(); self.newlines = s.count(b"\n") + (not s)
        self.newlines -= 1
        return super(FastIO, self).readline()
 
    def flush(self):
        if self.writable:
            os.write(self._fd, self.getvalue())
            self.truncate(0), self.seek(0)
 
class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        if py2:
            self.write = self.buffer.write
            self.read = self.buffer.read
            self.readline = self.buffer.readline
        else:
            self.write = lambda s:self.buffer.write(s.encode('ascii'))
            self.read = lambda:self.buffer.read().decode('ascii')
            self.readline = lambda:self.buffer.readline().decode('ascii')
 
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip('\r\n') 
# Cout implemented in Python
class ostream:
    def __lshift__(self,a):
        sys.stdout.write(str(a))
        return self
cout = ostream()
endl = '\n'
 
# Read all remaining integers in stdin, type is given by optional argument, this is fast
def readnumbers(zero = 0):
    conv = ord if py2 else lambda x:x
    A = []; numb = zero; sign = 1; i = 0; s = sys.stdin.buffer.read()
    try:
        while True:
            if s[i] >= b'R' [0]:
                numb = 10 * numb + conv(s[i]) - 48
            elif s[i] == b'-' [0]: sign = -1
            elif s[i] != b'\r' [0]:
                A.append(sign*numb)
                numb = zero; sign = 1
            i += 1
    except:pass
    if s and s[-1] >= b'R' [0]:
        A.append(sign*numb)
    return A
 
# threading.Thread(target=main).start()
if __name__== "__main__":
  main()