Problem: 2^k teams participate in a playoff tournament. The tournament consists of 2^k - 1 games. They are held as follows: first of all, the teams are split into pairs: team 1 plays against team 2, team 3 plays against team 4 (exactly in this order), and so on (so, 2^{k-1} games are played in that phase). When a team loses a game, it is eliminated, and each game results in elimination of one team (there are no ties). After that, only 2^{k-1} teams remain. If only one team remains, it is declared the champion; otherwise, 2^{k-2} games are played: in the first one of them, the winner of the game "1 vs 2" plays against the winner of the game "3 vs 4", then the winner of the game "5 vs 6" plays against the winner of the game "7 vs 8", and so on. This process repeats until only one team remains.

For example, this picture describes the chronological order of games with k = 3:

<image>

Let the string s consisting of 2^k - 1 characters describe the results of the games in chronological order as follows:

  * if s_i is 0, then the team with lower index wins the i-th game; 
  * if s_i is 1, then the team with greater index wins the i-th game; 
  * if s_i is ?, then the result of the i-th game is unknown (any team could win this game). 



Let f(s) be the number of possible winners of the tournament described by the string s. A team i is a possible winner of the tournament if it is possible to replace every ? with either 1 or 0 in such a way that team i is the champion.

You are given the initial state of the string s. You have to process q queries of the following form: 

  * p c — replace s_p with character c, and print f(s) as the result of the query. 

Input

The first line contains one integer k (1 ≤ k ≤ 18).

The second line contains a string consisting of 2^k - 1 characters — the initial state of the string s. Each character is either ?, 0, or 1.

The third line contains one integer q (1 ≤ q ≤ 2 ⋅ 10^5) — the number of queries.

Then q lines follow, the i-th line contains an integer p and a character c (1 ≤ p ≤ 2^k - 1; c is either ?, 0, or 1), describing the i-th query.

Output

For each query, print one integer — f(s).

Example

Input


3
0110?11
6
5 1
6 ?
7 ?
1 ?
5 ?
1 1


Output


1
2
3
3
5
4

Solution (in PYTHON3):

Solution:
import sys
read=sys.stdin.buffer.read;readline=sys.stdin.buffer.readline;input=lambda:sys.stdin.readline().rstrip()
import bisect,string,math,time,functools,random,fractions
from bisect import*
from heapq import heappush,heappop,heapify
from collections import deque,defaultdict,Counter
from itertools import permutations,combinations,groupby
rep=range;R=range
def I():return int(input())
def LI():return [int(i) for i in input().split()]
def LI_():return [int(i)-1 for i in input().split()]
def AI():return map(int,open(0).read().split())
def S_():return input()
def IS():return input().split()
def LS():return [i for i in input().split()]
def NI(n):return [int(input()) for i in range(n)]
def NI_(n):return [int(input())-1 for i in range(n)]
def NLI(n):return [[int(i) for i in input().split()] for i in range(n)]
def NLI_(n):return [[int(i)-1 for i in input().split()] for i in range(n)]
def StoLI():return [ord(i)-97 for i in input()]
def ItoS(n):return chr(n+97)
def LtoS(ls):return ''.join([chr(i+97) for i in ls])
def RLI(n=8,a=1,b=10):return [random.randint(a,b)for i in range(n)]
def RI(a=1,b=10):return random.randint(a,b)
def INP():
    N=10
    n=random.randint(1,N)
    a=RLI(n,0,n-1)
    return n,a
def Rtest(T):
    case,err=0,0
    for i in range(T):
        inp=INP()
        a1=naive(*inp)
        a2=solve(*inp)
        if a1!=a2:
            print(inp)
            print('naive',a1)
            print('solve',a2)
            err+=1
        case+=1
    print('Tested',case,'case with',err,'errors')
def GI(V,E,ls=None,Directed=False,index=1):
    org_inp=[];g=[[] for i in range(V)]
    FromStdin=True if ls==None else False
    for i in range(E):
        if FromStdin:
            inp=LI()
            org_inp.append(inp)
        else:
            inp=ls[i]
        if len(inp)==2:a,b=inp;c=1
        else:a,b,c=inp
        if index==1:a-=1;b-=1
        aa=(a,c);bb=(b,c);g[a].append(bb)
        if not Directed:g[b].append(aa)
    return g,org_inp
def GGI(h,w,search=None,replacement_of_found='.',mp_def={'#':1,'.':0},boundary=1):
    #h,w,g,sg=GGI(h,w,search=['S','G'],replacement_of_found='.',mp_def={'#':1,'.':0},boundary=1) # sample usage
    mp=[boundary]*(w+2);found={}
    for i in R(h):
        s=input()
        for char in search:
            if char in s:
                found[char]=((i+1)*(w+2)+s.index(char)+1)
                mp_def[char]=mp_def[replacement_of_found]
        mp+=[boundary]+[mp_def[j] for j in s]+[boundary]
    mp+=[boundary]*(w+2)
    return h+2,w+2,mp,found
def TI(n):return GI(n,n-1)
def accum(ls):
    rt=[0]
    for i in ls:rt+=[rt[-1]+i]
    return rt
def bit_combination(n,base=2):
    rt=[]
    for tb in R(base**n):s=[tb//(base**bt)%base for bt in R(n)];rt+=[s]
    return rt
def gcd(x,y):
    if y==0:return x
    if x%y==0:return y
    while x%y!=0:x,y=y,x%y
    return y
def YN(x):print(['NO','YES'][x])
def Yn(x):print(['No','Yes'][x])
def show(*inp,end='\n'):
    if show_flg:print(*inp,end=end)

mo=10**9+7
#mo=998244353
inf=float('inf')
FourNb=[(-1,0),(1,0),(0,1),(0,-1)];EightNb=[(-1,0),(1,0),(0,1),(0,-1),(1,1),(-1,-1),(1,-1),(-1,1)];compas=dict(zip('WENS',FourNb));cursol=dict(zip('LRUD',FourNb))
alp=[chr(ord('a')+i)for i in range(26)]
#sys.setrecursionlimit(10**7)
def gcj(c,x):
    print("Case #{0}:".format(c+1),x)

show_flg=False
show_flg=True

## Segment Tree ##

## Test case: ABC 146 F
## https://atcoder.jp/contests/abc146/tasks/abc146_f

## Initializer Template ##
# Range Sum:        sg=SegTree(n)
# Range Minimum:    sg=SegTree(n,inf,min,inf)

class SegTree:
    def __init__(self,n,ls,init_val=1,function=lambda a,b:a+b,ide=0):
        self.size=n
        self.ls=ls
        self.ide_ele=ide
        self.num=1<<(self.size-1).bit_length()
        self.table=[self.ide_ele]*2*self.num
        self.index=[0]*2*self.num
        
        def function(a,b,k):
            if self.ls[k]==1:
                res=self.table[k*2+1]
            elif self.ls[k]==0:
                res=self.table[k*2+2]
            else:
                res=self.table[k*2+1]+self.table[k*2+2]
            return res
        self.func=function
        #set_val
        if not hasattr(init_val,"__iter__"):
            init_val=[init_val]*self.size
        for i,val in enumerate(init_val):
            self.table[i+self.num-1]=val
            self.index[i+self.num-1]=i
        #build
        for i in range(self.num-2,-1,-1):
            self.table[i]=self.func(self.table[2*i+1],self.table[2*i+2],i)
            if self.table[i]==self.table[i*2+1]:
                self.index[i]=self.index[i*2+1]
            else:
                self.index[i]=self.index[i*2+2]
    def build(self):
        for i in range(self.num-2,-1,-1):
            self.table[i]=self.func(self.table[2*i+1],self.table[2*i+2],i)
            if self.table[i]==self.table[i*2+1]:
                self.index[i]=self.index[i*2+1]
            else:
                self.index[i]=self.index[i*2+2]
    
    def update(self,k,x):
        self.ls[k]=x
        while k>-1:
            res=self.func(self.table[k*2+1],self.table[k*2+2],k)
            self.table[k]=res
            k=(k-1)//2
            

    def query(self):
        if q<=p:
            return self.ide_ele
        p+=self.num-1
        q+=self.num-2
        res=self.ide_ele
        while q-p>1:
            if p&1==0:
                res=self.func(res,self.table[p])
            if q&1==1:
                res=self.func(res,self.table[q])
                q-=1
            p=p>>1
            q=(q-1)>>1
        if p==q:
            res=self.func(res,self.table[p])
        else:
            res=self.func(self.func(res,self.table[p]),self.table[q])
        return res

    def __str__(self):
        # 生配列を表示
        rt=self.table[self.num-1:self.num-1+self.size]
        return str(rt)

n=int(input())
dc=dict(zip('01?',[0,1,2]))
s=[dc[i]for i in input()][::-1]

sg=SegTree(2**n,s)

y=2**n-1

for i in range(int(input())):
    x,c=input().split()
    x=int(x)-1
    c=dc[c]
    sg.update(y-x-1,c)
    ans=sg.table[0]
    
    print(ans)