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 CPP):

Solution:
#include<bits/stdc++.h>
#define ll long long 
#define int ll
#define pb push_back
#define ff first
#define endl "\n"
#define ss second
#define setbits(x)          __builtin_popcountll(x)
#define lp(x,a,n) for(ll x=a; x<n; x++)
#define lpi(x,a,n) for(int x=a; x<n; x++)
#define w(t) int t; cin>>t; while(t--)
#define vi vector<int> 
#define vll vector<ll>
#define pll pair<ll, ll>
#define pii pair<int, int>
#define inf 1e9
#define LB lower_bound
#define UB upper_bound
const ll mod = 1000000007;
using namespace std;
ll gcd(ll a,ll b){
    if(a==0)return b;
    return gcd(b%a,a);
}
ll lcm(ll a, ll b){
    return (a*b)/gcd(a,b);
}
 
 
// use when we need to find (a/b)%mod value, where a%mod already happened.
// ex we need to find ans/2 and ans already is in the form of ans%mod
// then we can use this statement.
// ans *= find(2, mod-2, mod);
// ans %=  mod;
 
ll divideBymod(ll x, ll y, ll p)
{   ll res = 1; x = x % p;
    while (y > 0)
    {if(y & 1) res = (res*x) % p;
    y = y>>1;x = (x*x) % p;
    }return res;
}
 
int prime[1007];
void SieveOfEratosthenes(int n) 
{ 
    memset(prime, -1, sizeof(prime)); 
  
    for (int p=2; p*p<=n; p++) 
    { 
         
        if (prime[p] == -1) 
        { 
            for (int i=p*p; i<=n; i += p) 
                prime[i] = p; 
        } 
    } 
    
} 


ll power(ll x, ll y) 
{ 
    ll res = 1;     
    while (y > 0) 
    { 
        if (y & 1) 
            res = (res*x)%mod; 
   
        y = y>>1;
        x = (x*x)%mod;  
    } 
    return res; 
}
 

ll fact[3], invfact[3];

void factorial(ll n){
    fact[0] = 1;
    for(ll i=1; i<n; i++){
        fact[i] = (fact[i-1]*i)%mod;
    }
    invfact[n-1] = power(fact[n-1], mod-2ll); //here we are passing mod-2 as power and mod to mod the ans generated
    for(int i=n-2; i>=0; i--){
        invfact[i] = (invfact[i+1]*(i+1))%mod;
    }
}



int LPS(string s, int dp[]){
    int len = 0;
    int n = s.length();
    dp[0] = 0;
    for(int i=1; i<n; i++){
        if(s[i] == s[len]){
            len++;
            dp[i] = len;
            continue;
        }
        if(len!=0){
            len = dp[len-1];
            i--;
            continue;
        }
        dp[i]=0;
    }
    return dp[n-1];
}

bool KMPSearch(string pat, string txt)
{
    int M = pat.length();
    int N = txt.length();
  
    int lps[M];
    LPS(pat, lps);
  
    int i = 0; 
    int j = 0; 
    while (i < N) {
        if (pat[j] == txt[i]) {
            j++;
            i++;
        }
  
        if (j == M) {
            return true;
            j = lps[j - 1];
        }
        else if (i < N && pat[j] != txt[i]) {
            if (j != 0)
                j = lps[j - 1];
            else
                i = i + 1;
        }
    }
    return false;
}
 

//to calculate p1^x  *  p2^x2  *p3^x3 ----------
//i.e. prime factorization and store pi and xi in vector 
vector<pair<ll, ll>> pFactor;
 
void factorize(long long n) 
{ 
    int count = 0; 
  
    // count the number of times 2 divides  
    while (!(n % 2)) { 
        n >>= 1; // equivalent to n=n/2; 
        count++; 
    } 
  
    // if 2 divides it 
    if (count) 
        pFactor.pb({2,count}); 
  
    // check for all the possible numbers that can  
    // divide it 
    for (long long i = 3; i <= sqrt(n); i += 2) { 
        count = 0; 
        while (n % i == 0) { 
            count++; 
            n = n / i; 
        } 
        if (count) 
            pFactor.pb({i,count});
    } 
  
    // if n at the end is a prime number. 
    if (n > 2) 
        pFactor.pb({n,1}); 
} 
 

 
int32_t main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    //SieveOfEratosthenes(10000007);
    
        int k;
        cin>>k;
        string s;
        cin>>s;
        int n=s.length();
        int adj[n+1];
        int revadj[n+1][2];
        int j=n;
        int i=n-1;
        while(i>0){
            adj[i] = j;
            revadj[j][1] = i;
            i--;
            
            adj[i] = j;
            revadj[j][0] = i;
            i--;
            j--;
        }
        int val[n+1];
        lp(i,1, n+1){
            if(i<=((1<<k)/2)){
                if(s[i-1]=='?'){
                    val[i] = 2;
                }else {
                    val[i] = 1;
                }
            }else{
                int right = revadj[i][1], left = revadj[i][0];
                right = val[right];
                left = val[left];
                if(s[i-1]=='?'){
                    val[i] = left+right;
                }else if(s[i-1]=='1'){
                    val[i] = right;
                }else val[i]=left;
            }
        }
        int q;
        cin>>q;
        while(q--){
            int p;
            char c;
            cin>>p>>c;
            int updt=0;
            if(p<=((1<<k)/2)){
                if(c=='?'){
                    if(s[p-1]=='?') {
                        goto there;
                    }
                    updt=2;
                }else {
                    if(s[p-1]=='?'){
                        updt=1;
                    }else {
                        goto there;
                    }
                }
            }else {
                int left = revadj[p][0], right= revadj[p][1];
                left = val[left];
                right = val[right];
                
                if(c=='?'){
                    if(s[p-1]=='?'){
                        goto there;
                    }else {
                        updt = left+right;
                    }
                }else {
                    if(c=='0'){
                        updt = left;
                    }else {
                        updt = right;
                    }
                }
            }
            s[p-1] = c;
            val[p] = updt;
            while(p!=n){
                int u = adj[p];
                if(s[u-1]=='1'){
                    if(revadj[u][0]==p){
                        break;
                    }else {
                        val[u] = val[p];
                    }
                }else if(s[u-1]=='0'){
                    if(revadj[u][0]==p){
                        val[u] = val[p];
                    }else break;
                }else {
                    val[u] = val[revadj[u][0]] + val[revadj[u][1]];
                }
                p = u;
            }
            there:
            cout<<val[n]<<endl;
        }
    

}