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:
#!/usr/bin/env python3
import sys
import getpass  # not available on codechef
import math, random
import functools, itertools, collections, heapq, bisect
from collections import Counter, defaultdict, deque
input = sys.stdin.readline  # to read input quickly

# available on Google, AtCoder Python3, not available on Codeforces
# import numpy as np
# import scipy

M9 = 10**9 + 7  # 998244353
yes, no = "YES", "NO"
# d4 = [(1,0),(0,1),(-1,0),(0,-1)]
# d8 = [(1,0),(1,1),(0,1),(-1,1),(-1,0),(-1,-1),(0,-1),(1,-1)]
# d6 = [(2,0),(1,1),(-1,1),(-2,0),(-1,-1),(1,-1)]  # hexagonal layout
MAXINT = sys.maxsize

# if testing locally, print to terminal with a different color
OFFLINE_TEST = getpass.getuser() == "hkmac"
# OFFLINE_TEST = False  # codechef does not allow getpass
def log(*args):
    if OFFLINE_TEST:
        print('\033[36m', *args, '\033[0m', file=sys.stderr)

def solve(*args):
    # screen input
    if OFFLINE_TEST:
        log("----- solving ------")
        log(*args)
        log("----- ------- ------")
    return solve_(*args)

def read_matrix(rows):
    return [list(map(int,input().split())) for _ in range(rows)]

def read_strings(rows):
    return [input().strip() for _ in range(rows)]

def minus_one(arr):
    return [x-1 for x in arr]

def minus_one_matrix(mrr):
    return [[x-1 for x in row] for row in mrr]

# ---------------------------- template ends here ----------------------------


def solve_(srr, qrr, k):
    # your solution here

    mappping = {"?": 2, "1": 1, "0":0}
    status = [mappping[x] for x in srr]
    counts = [1 for _ in srr]  # result at -1
    nextidx = [-1 for _ in srr]
    prevsmall_arr = [-1 for _ in srr]
    prevlarge_arr = [-1 for _ in srr]

    idx = 0
    
    for j in range(k-1,-1,-1):
        prevsize = 2**(j+1)
        startidx = idx
        for i in range(2**j):
            if j == k-1:
                if status[idx] == 2:
                    counts[idx] = 2
            else:
                prevsmall = startidx - prevsize + (idx - startidx)*2
                prevlarge = prevsmall + 1
                nextidx[prevsmall] = idx
                nextidx[prevlarge] = idx
                prevsmall_arr[idx] = prevsmall
                prevlarge_arr[idx] = prevlarge
                # log(idx, prevsmall, prevlarge)

                if status[idx] == 0:
                    counts[idx] = counts[prevsmall]
                if status[idx] == 1:
                    counts[idx] = counts[prevlarge]
                if status[idx] == 2:
                    counts[idx] = counts[prevsmall] + counts[prevlarge]
                
            idx += 1

    # log("chec")
    # log(prevsmall_arr)
    # log(prevlarge_arr)
    # log(nextidx)

    res = []
    for row in qrr:
        idx, op = row.split()
        idx = int(idx)-1
        op = mappping[op]

        status[idx] = op

        if idx < 2**(k-1):
            # log("smaller than 2**k-1", 2**(k-1))
            if status[idx] == 0:
                counts[idx] = 1
            if status[idx] == 1:
                counts[idx] = 1
            if status[idx] == 2:
                counts[idx] = 2
        else:
            prevsmall = prevsmall_arr[idx]
            prevlarge = prevlarge_arr[idx]
            if status[idx] == 0:
                counts[idx] = counts[prevsmall]
            if status[idx] == 1:
                counts[idx] = counts[prevlarge]
            if status[idx] == 2:
                counts[idx] = counts[prevsmall] + counts[prevlarge]

        while nextidx[idx] != -1:
            idx = nextidx[idx]
            prevsmall = prevsmall_arr[idx]
            prevlarge = prevlarge_arr[idx]
            if status[idx] == 0:
                counts[idx] = counts[prevsmall]
            if status[idx] == 1:
                counts[idx] = counts[prevlarge]
            if status[idx] == 2:
                counts[idx] = counts[prevsmall] + counts[prevlarge]
        # log("-")
        # log(status)
        # log(counts)
        res.append(counts[-1])
    
    # log(res)
    return res


for case_num in [0]:  # no loop over test case
# for case_num in range(100):  # if the number of test cases is specified
# for case_num in range(int(input())):

    # read line as an integer
    k = int(input())

    # read line as a string
    srr = input().strip()

    q = int(input())

    # read one line and parse each word as a string
    # lst = input().split()
    
    # read one line and parse each word as an integer
    # a,b,c = list(map(int,input().split()))
    # lst = list(map(int,input().split()))
    # lst = minus_one(lst)

    # read multiple rows
    qrr = read_strings(q)  # and return as a list of str
    # mrr = read_matrix(k)  # and return as a list of list of int
    # mrr = minus_one_matrix(mrr)

    res = solve(srr, qrr, k)  # include input here

    # print length if applicable
    # print(len(res))

    # parse result
    # res = " ".join(str(x) for x in res)
    res = "\n".join(str(x) for x in res)
    # res = "\n".join(" ".join(str(x) for x in row) for row in res)

    # print result
    # print("Case #{}: {}".format(case_num+1, res))   # Google and Facebook - case number required

    print(res)